How To Evaluate Limits

Quick Reference

Indeterminate Form	Recommended Technique	Key Algebraic Operation
Direct Form $\frac{P (a)}{Q (a)}$ ( $Q (a) \neq 0$ )	Direct Substitution	Substitute $x = a$ directly into expression.
$\frac{0}{0}$ (Polynomials)	Factoring Technique	Factor out $(x - a)$ from numerator and denominator and cancel.
$\frac{0}{0}$ (Radicals)	Rationalizing Technique	Multiply numerator and denominator by conjugate radical $A + B \sqrt{C}$ .
$\frac{0}{0}$ or $\infty - \infty$ (Fractions)	Common Denominator Technique	Combine sub-fractions over a common denominator into a single fraction.
$\frac{\infty}{\infty}$ or Polynomials ( $x \to \pm \infty$ )	Leading Term Technique	Keep only the largest degree term of numerator and denominator.

To evaluate the limit of a function f as $x \to a$ , we first try putting x = a in $f (x)$ , and that is the value of the limit unless we get one of the indeterminate forms ( $0 / 0$ , $\pm \infty / \pm \infty$ , $0 \cdot (\pm \infty)$ , $\infty - \infty$ ). If we get an indeterminate form, to evaluate the limit, we can try one of the following strategies. The most powerful technique for finding limits of indeterminate forms will be discussed in L'Hôpital's Rule.

Note: Direct Substitution Works Only for Continuous Functions

This technique works when f is continuous at a, since $lim_{x \to a} f (x) = f (a)$ . However, we must exercise caution when applying direct substitution to piecewise-defined functions or at any point where the function may be discontinuous. For example, consider the greatest integer function (also known as the floor function), $f (x) = [[x]]$ . Although $[[2]] = 2$ , the limit $lim_{x \to 2} [[x]]$ does not exist because the one-sided limits differ: $lim_{x \to 2^{+}} [[x]]$ while $lim_{x \to 2^{-}} [[x]]$ (see the graph below).

Graph of floor function — Graph of $f (x) = [[x]]$ .

Factoring technique: If $P (x)$ and $Q (x)$ are two polynomial functions and $P (a) = Q (a) = 0$ , it follows that $(x - a)$ is a common factor (by the Factor Theorem) of $P (x)$ and $Q (x)$ . To evaluate $lim_{x \to a} \frac{P (x)}{Q (x)}$ , factor out $(x - a)$ and cancel it from the denominator and numerator.

If it is not easy to factor out $(x - a)$ , you may divide both $P (x)$ and $Q (x)$ by $x - a$ .

Find $lim_{x \to 2} \frac{x^{2} - 4}{3 x - 6}$ .

Solution

Plugging $x = 2$ into the expression $(x^{2} - 4) / (3 x - 6)$ yields $\frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work. The fact that both $x^{2} - 4$ and $3 x - 6$ become zero upon substituting $2$ for x shows that $x - 2$ is a common factor:

x^{2} - 4 = (x - 2) (x + 2)

3 x - 6 = 3 (x - 2) .

Therefore, we can simplify the fraction:

\begin{aligned} \lim_{x\to 2} \frac{x^2 - 4}{3x - 6} &= \lim_{x\to 2} \frac{(x - 2)(x + 2)}{3(x - 2)} \\ &= \frac{1}{3} \lim_{x\to 2} (x + 2). \end{aligned}

Now we can substitute $2$ for x:

\begin{aligned} \frac{1}{3} \lim_{x\to 2} (x + 2) &= \frac{1}{3}(2 + 2) \\ &= \frac{4}{3}. \end{aligned}

Find $lim_{x \to 1} \frac{x^{3} - 1}{2 x^{2} + x - 3}$ .

Solution

If we plug $x = 1$ into the given expression, we get $\frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work. Because $x^{3} - 1 = x^{3} - 1^{3}$ , we can use the Difference of Cubes formula $A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})$ and write:

x^{3} - 1^{3} = (x - 1) (x^{2} + x + 1) .

The denominator is a quadratic function, and we know $x - 1$ is a factor. The other factor is of the form $a x + b$ . To discover a and b, we multiply it by $x - 1$ :

(x - 1) (a x + b) = a x^{2} + (b - a) x - b .

By comparing the above expression with $2 x^{2} + x - 3$ , we realize that we must have $a = 2$ and $b = 3$ . Therefore,

2 x^{2} + x - 3 = (x - 1) (2 x + 3),

and

\begin{aligned} \frac{x^3 - 1}{2x^2 + x - 3} &= \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(2x + 3)} \\ &= \frac{x^2 + x + 1}{2x + 3} \qquad (\text{if } x \neq 1). \end{aligned}

Now we can plug $x = 1$ into the simplified fraction:

\begin{aligned} \lim_{x\to 1} \frac{x^2 + x + 1}{2x + 3} &= \frac{1^2 + 1 + 1}{2(1) + 3} \\ &= \frac{3}{5}. \end{aligned}

Rationalizing Technique: If there is an indeterminate limit, and the denominator or the numerator is an expression of the form $A - B \sqrt{C}$ which becomes $0$ upon substitution of a for x, multiply both the numerator and denominator by the conjugate radical $A + B \sqrt{C}$ and use the following formula:

(A - B \sqrt{C}) (A + B \sqrt{C}) = A^{2} - B^{2} C . (Recall (a - b) (a + b) = a^{2} - b^{2})

Find $lim_{x \to - 4} \frac{x^{2} - 16}{3 - \sqrt{x^{2} - 7}}$ .

Solution

If we substitute $- 4$ for x, we get the indeterminate form $\frac{0}{0}$ , so direct substitution does not work. As discussed above, we multiply and divide the given expression by the conjugate of the denominator, which is $3 + \sqrt{x^{2} - 7}$ :

\begin{aligned} \lim_{x\to -4} \frac{x^2 - 16}{3 - \sqrt{x^2 - 7}} &= \lim_{x\to -4} \left( \frac{x^2 - 16}{3 - \sqrt{x^2 - 7}} \cdot \frac{3 + \sqrt{x^2 - 7}}{3 + \sqrt{x^2 - 7}} \right) \\ &= \lim_{x\to -4} \frac{(x^2 - 16)(3 + \sqrt{x^2 - 7})}{3^2 - (x^2 - 7)} \\ &= \lim_{x\to -4} \frac{(x^2 - 16)(3 + \sqrt{x^2 - 7})}{16 - x^2} \\ &= \lim_{x\to -4} \left( -3 - \sqrt{x^2 - 7} \right) \\ &= -3 - 3 \\ &= -6. \end{aligned}

Common Denominator Technique: If the limit of $f \pm g$ is indeterminate and f and g are two fractions, find a common denominator, convert them to one indeterminate quotient ( $0 / 0$ or $\pm \infty / \pm \infty$ ), and then simplify the result.

Find $lim_{x \to 1} (\frac{2}{x^{2} - 1} - \frac{1}{x - 1})$ .

Solution

When $x \to 1$ and $x > 1$ (that is, as $x \to 1^{+}$ ), both $x^{2} - 1$ and $x - 1$ approach zero through positive values. So

lim_{x \to 1^{+}} \frac{2}{x^{2} - 1} \overset{[\frac{2}{0^{+}}]}{=} + \infty, lim_{x \to 1^{+}} \frac{1}{x - 1} \overset{[\frac{1}{0^{+}}]}{=} + \infty

(Note: We write $\frac{2}{0^{+}}$ on the equal sign to denote that the limit of the numerator is $2$ and the limit of the denominator is zero, but the denominator approaches through positive values. In other words, the numerator and the denominator are of the same sign. Therefore, the fraction goes to $+ \infty$ , not $- \infty$ ) and we have an indeterminate form $\infty - \infty$ , so direct evaluation does not work. When $x \to 1$ and $x < 1$ (that is, as $x \to 1^{-}$ ), then both $x^{2} - 1$ and $x - 1$ approach zero through negative values. So

lim_{x \to 1^{-}} \frac{2}{x^{2} - 1} \overset{[\frac{2}{0^{-}}]}{=} - \infty, lim_{x \to 1^{-}} \frac{1}{x - 1} \overset{[\frac{1}{0^{-}}]}{=} - \infty

and we again have an indeterminate form $- \infty - (- \infty)$ or $- \infty + \infty$ .

To evaluate the limit, we find the common denominator. Because $x^{2} - 1 = (x - 1) (x + 1)$ [Recall the identity $A^{2} - B^{2} = (A - B) (A + B)$ ], we have

\begin{aligned} \lim_{x\to 1} \left( \frac{2}{x^2 - 1} - \frac{1}{x - 1} \right) &= \lim_{x\to 1} \left( \frac{2}{(x - 1)(x + 1)} - \frac{x + 1}{(x - 1)(x + 1)} \right) \\ &= \lim_{x\to 1} \frac{2 - (x + 1)}{(x - 1)(x + 1)} \\ &= \lim_{x\to 1} \frac{1 - x}{(x - 1)(x + 1)} \\ &= \lim_{x\to 1} \frac{-1}{x + 1} \\ &= -\frac{1}{2}. \end{aligned}

Here, we did not have to consider the one-sided limits ( $x \to 1^{+}$ and $x \to 1^{-}$ ) separately.

Leading Term Technique: When x is numerically large ( $x \to + \infty$ or $x \to - \infty$ ), the limit of a polynomial and a rational function is determined by their leading terms as explained below.

Find $lim_{x \to - \infty} (- 3 x^{5} - 8 x^{2} + 37)$ .

Solution

Because $- 3 x^{5} \to (- 3) (- \infty) = + \infty$ and $- 8 x^{2} \to - 8 (+ \infty) = - \infty$ , the limit has the indeterminate form $\infty - \infty$ , so direct evaluation does not work. To evaluate this limit, we factor out the highest power of x and then use the fact that $lim_{x \to \pm \infty} \frac{1}{x^{r}} = 0$ for $r > 0$ :

\begin{aligned} \lim_{x\to -\infty} (-3x^5 - 8x^2 + 37) &= \lim_{x\to -\infty} \left[ x^5 \left( -3 - \frac{8}{x^3} + \frac{37}{x^5} \right) \right] \\ &= \left( \lim_{x\to -\infty} x^5 \right) \left( \lim_{x\to -\infty} \left( -3 - \frac{8}{x^3} + \frac{37}{x^5} \right) \right) \\ &= \left( \lim_{x\to -\infty} x^5 \right) (-3 - 0 + 0) \\ &= (-\infty)(-3) \\ &= +\infty. \end{aligned}

Similar to the last example, as $x \to + \infty$ or $x \to - \infty$ , we can factor out the highest power of x and show that if $a_{n} \neq 0$ then:

\bbox[8px,#E6F0FA, border: 3px solid #0066CC]{ \begin{aligned} \lim_{x\to\pm\infty} &(a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0) = \lim_{x\to\pm\infty} (a_n x^n) \\ &= \text{limit of the largest degree term.} \end{aligned} }

Similarly, to determine the behavior of a rational function (a function of the form $\frac{P (x)}{Q (x)}$ where $P (x)$ and $Q (x)$ are polynomials) as $x \to + \infty$ or $x \to - \infty$ , we first divide each term in the numerator by the highest power of x occurring in the numerator and each term in the denominator by the highest power of x occurring in the denominator, and then use the fact that $\frac{1}{x^{r}} \to 0$ ( $r > 0$ ) as $x \to \pm \infty$ . It follows from this argument that if $a_{n} \neq 0$ and $b_{m} \neq 0$ , then:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\begin{aligned} \lim_{x\to\pm\infty} &\frac{a_n x^n + \dots + a_1 x + a_0}{b_m x^m + \dots + b_1 x + b_0} \\ &= \lim_{x\to\pm\infty} \frac{a_n x^n}{b_m x^m} \\ &= \lim_{x\to\pm\infty} \frac{\text{largest degree term of numerator}}{\text{largest degree term of denominator}}. \end{aligned}}

Find:

$lim_{x \to - \infty} \frac{3 x^{3} - 4 x^{2} + x + 3}{- x^{3} + 2 x^{2}}$
$lim_{x \to + \infty} \frac{13 x^{4} - 5 x^{3} + 4 x + 15}{- x^{5} + 16 x^{2} + 1}$

Solution

(a) Direct evaluation as $x \to - \infty$ gives $\frac{\infty}{\infty}$ , which is an indeterminate form, so direct evaluation does not work. Let's factor out the highest power of x and then use the fact that $\frac{1}{x^{r}} \to 0$ ( $r > 0$ ) as $x \to \pm \infty$ :

\begin{aligned} \lim_{x\to -\infty} \frac{3x^3 - 4x^2 + x + 3}{-x^3 + 2x^2} &= \lim_{x\to -\infty} \frac{x^3 \left( 3 - \frac{4}{x} + \frac{1}{x^2} + \frac{3}{x^3} \right)}{-x^3 \left( 1 - \frac{2}{x} \right)} \\ &= \frac{\lim_{x\to -\infty} 3 - 4 \lim_{x\to -\infty} \frac{1}{x} + \lim_{x\to -\infty} \frac{1}{x^2} + 3 \lim_{x\to -\infty} \frac{1}{x^3}}{\lim_{x\to -\infty} (-1) - 2 \lim_{x\to -\infty} \frac{1}{x}} \\ &= \frac{3 - 0 + 0 + 0}{-1 + 0} \\ &= -3, \end{aligned}

or we may use the above shortcut and say:

\begin{aligned} \lim_{x\to -\infty} \frac{3x^3 - 4x^2 + x + 3}{-x^3 + 2x^2} &= \lim_{x\to -\infty} \frac{3x^3}{-x^3} \\ &= -3. \end{aligned}
(b) Direct evaluation as

x \to + \infty

gives

\frac{\infty}{\infty}

, so direct evaluation does not work. We use the leading term shortcut: to determine the end behavior of a rational function, we keep only the largest degree term in the numerator and the largest degree term in the denominator: \begin{aligned} \lim_{x\to +\infty} \frac{13x^4 - 5x^3 + 4x + 15}{-x^5 + 16x^2 + 1} &= \lim_{x\to +\infty} \frac{13x^4}{-x^5} \\ &= \lim_{x\to +\infty} \frac{-13}{x} \\ &= 0. \end{aligned}

Exercises

Evaluate the limit or show that it does not exist:

lim_{t \to 2} \frac{t^{2} + 5 t + 6}{t + 2}

Answer

$5$

Solution

By direct substitution of $t = 2$ :

\begin{aligned} \lim_{t\to 2} \frac{t^2 + 5t + 6}{t + 2} &= \frac{2^2 + 5(2) + 6}{2 + 2} \\ &= \frac{4 + 10 + 6}{4} \\ &= \frac{20}{4} \\ &= 5. \end{aligned}

Since direct substitution yields a defined value ( $5$ ), this is not an indeterminate form.

Evaluate the limit or show that it does not exist:

lim_{t \to - 2} \frac{t^{2} + 5 t + 6}{t + 2}

Answer

$1$

Solution

Direct substitution of $t = - 2$ gives $\frac{(- 2)^{2} + 5 (- 2) + 6}{- 2 + 2} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

To factor the numerator $t^{2} + 5 t + 6$ , we need two numbers whose sum is $5$ and whose product is $6$ . These numbers are $2$ and $3$ , so $t^{2} + 5 t + 6 = (t + 2) (t + 3)$ .

Canceling the common factor $(t + 2)$ for $t \neq - 2$ :

\begin{aligned} \lim_{t\to -2} \frac{t^2 + 5t + 6}{t + 2} &= \lim_{t\to -2} \frac{(t + 2)(t + 3)}{t + 2} \\ &= \lim_{t\to -2} (t + 3) \\ &= -2 + 3 \\ &= 1. \end{aligned}

Evaluate the limit or show that it does not exist:

lim_{y \to 5} \frac{y - 5}{y^{2} - 25}

Answer

$\frac{1}{10}$

Solution

Direct substitution of $y = 5$ gives $\frac{5 - 5}{5^{2} - 25} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Factoring the denominator using the difference of squares:

y^{2} - 25 = (y - 5) (y + 5) .

Canceling the common factor $(y - 5)$ for $y \neq 5$ :

\begin{aligned} \lim_{y\to 5} \frac{y - 5}{y^2 - 25} &= \lim_{y\to 5} \frac{y - 5}{(y - 5)(y + 5)} \\ &= \lim_{y\to 5} \frac{1}{y + 5} \\ &= \frac{1}{5 + 5} \\ &= \frac{1}{10}. \end{aligned}

Evaluate the limit:

lim_{h \to 0} \frac{(x + h)^{2} - x^{2}}{h}

Answer

$2 x$

Solution

Direct substitution of $h = 0$ gives $\frac{(x + 0)^{2} - x^{2}}{0} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Expand the squared term in the numerator:

\begin{aligned} (x + h)^2 - x^2 &= x^2 + 2xh + h^2 - x^2 \\ &= 2xh + h^2 \\ &= h(2x + h). \end{aligned}

Canceling h for $h \neq 0$ :

\begin{aligned} \lim_{h\to 0} \frac{(x + h)^2 - x^2}{h} &= \lim_{h\to 0} \frac{h(2x + h)}{h} \\ &= \lim_{h\to 0} (2x + h) \\ &= 2x + 0 \\ &= 2x. \end{aligned}

Find $lim_{x \to 3} \frac{x^{3} - 27}{x^{3} - 2 x^{2} - 5 x + 6}$ .

Answer

$\frac{27}{10}$

Solution

Direct substitution of $x = 3$ gives $\frac{3^{3} - 27}{3^{3} - 2 (3)^{2} - 5 (3) + 6} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Using the Difference of Cubes formula $A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})$ :

x^{3} - 3^{3} = (x - 3) (x^{2} + 3 x + 9) .

Dividing denominator $x^{3} - 2 x^{2} - 5 x + 6$ by $(x - 3)$ gives:

x^{3} - 2 x^{2} - 5 x + 6 = (x - 3) (x^{2} + x - 2) .

Canceling $(x - 3)$ :

\begin{aligned} \lim_{x\to 3} \frac{x^3 - 27}{x^3 - 2x^2 - 5x + 6} &= \lim_{x\to 3} \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x^2 + x - 2)} \\ &= \lim_{x\to 3} \frac{x^2 + 3x + 9}{x^2 + x - 2} \\ &= \frac{3^2 + 3(3) + 9}{3^2 + 3 - 2} \\ &= \frac{27}{10}. \end{aligned}

Find $lim_{x \to 2} \frac{\frac{1}{x + 1} - \frac{1}{3}}{x - 2}$ .

Answer

$- \frac{1}{9}$

Solution

Direct substitution of $x = 2$ gives $\frac{\frac{1}{3} - \frac{1}{3}}{2 - 2} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Combine terms in numerator over a common denominator:

\begin{aligned} \lim_{x\to 2} \frac{\frac{1}{x+1} - \frac{1}{3}}{x - 2} &= \lim_{x\to 2} \frac{\frac{3 - (x + 1)}{3(x + 1)}}{x - 2} \\ &= \lim_{x\to 2} \frac{\frac{2 - x}{3(x + 1)}}{\frac{x - 2}{1}} \\ &= \lim_{x\to 2} \frac{2 - x}{3(x - 2)(x + 1)} \\ &= \lim_{x\to 2} \frac{-1}{3(x + 1)} \\ &= -\frac{1}{3(3)} \\ &= -\frac{1}{9}. \end{aligned}

Given $x > 0$ , find $lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$ .

Answer

$\frac{1}{2 \sqrt{x}}$

Solution

Direct substitution of $h = 0$ gives $\frac{\sqrt{x} - \sqrt{x}}{0} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Multiply numerator and denominator by the conjugate $\sqrt{x + h} + \sqrt{x}$ :

\begin{aligned} \lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} &= \lim_{h\to 0} \left( \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \right) \\ &= \lim_{h\to 0} \frac{(x + h) - x}{h(\sqrt{x+h} + \sqrt{x})} \\ &= \lim_{h\to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} \\ &= \lim_{h\to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} \\ &= \frac{1}{\sqrt{x} + \sqrt{x}} \\ &= \frac{1}{2\sqrt{x}}. \end{aligned}

Evaluate the limit:

lim_{x \to 0^{+}} \frac{\sqrt{x}}{\sqrt{9 + \sqrt{x}} - 3}

Answer

$6$

Solution

Direct substitution of $x = 0$ gives $\frac{0}{\sqrt{9} - 3} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Multiply numerator and denominator by the conjugate $\sqrt{9 + \sqrt{x}} + 3$ :

\begin{aligned} \lim_{x\to 0^+} \frac{\sqrt{x}}{\sqrt{9 + \sqrt{x}} - 3} &= \lim_{x\to 0^+} \frac{\sqrt{x} \left(\sqrt{9 + \sqrt{x}} + 3\right)}{\left(\sqrt{9 + \sqrt{x}} - 3\right)\left(\sqrt{9 + \sqrt{x}} + 3\right)} \\ &= \lim_{x\to 0^+} \frac{\sqrt{x} \left(\sqrt{9 + \sqrt{x}} + 3\right)}{(9 + \sqrt{x}) - 9} \\ &= \lim_{x\to 0^+} \frac{\sqrt{x} \left(\sqrt{9 + \sqrt{x}} + 3\right)}{\sqrt{x}} \\ &= \lim_{x\to 0^+} \left(\sqrt{9 + \sqrt{x}} + 3\right) \\ &= \sqrt{9 + 0} + 3 \\ &= 3 + 3 \\ &= 6. \end{aligned}

Evaluate the limit or show that it does not exist:

lim_{u \to 0} \frac{5 u^{3} + 9 u^{2}}{3 u^{4} - 12 u^{2}}

Answer

$- \frac{3}{4}$

Solution

Direct substitution of $u = 0$ gives $\frac{5 (0)^{3} + 9 (0)^{2}}{3 (0)^{4} - 12 (0)^{2}} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Factoring out $u^{2}$ from the numerator and denominator for $u \neq 0$ :

\begin{aligned} \frac{5u^3 + 9u^2}{3u^4 - 12u^2} &= \frac{u^2(5u + 9)}{u^2(3u^2 - 12)} \\ &= \frac{5u + 9}{3u^2 - 12}. \end{aligned}

Taking the limit as $u \to 0$ :

\begin{aligned} \lim_{u\to 0} \frac{5u + 9}{3u^2 - 12} &= \frac{5(0) + 9}{3(0)^2 - 12} \\ &= \frac{9}{-12} \\ &= -\frac{3}{4}. \end{aligned}

Evaluate the limit:

lim_{x \to 0^{+}} \frac{| x |}{x}

Answer

$1$

Solution

Direct substitution of $x = 0^{+}$ gives $\frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

As $x \to 0^{+}$ , $x > 0$ , so $| x | = x$ .

Thus:

\begin{aligned} \lim_{x\to 0^+} \frac{|x|}{x} &= \lim_{x\to 0^+} \frac{x}{x} \\ &= \lim_{x\to 0^+} 1 \\ &= 1. \end{aligned}

Evaluate the limit or show that it does not exist:

lim_{x \to 2^{+}} \frac{⌊ x ⌋^{2} - 4}{x^{2} - 4}

(Note: $⌊ x ⌋$ is the floor function, which is the same as the greatest integer function $[[x]]$ .)

Answer

$0$

Solution

Direct substitution is not applicable here in the usual sense because the floor function is piecewise constant on intervals. We do not have an indeterminate form $\frac{0}{0}$ ; rather, we are evaluating the limit of a constant function.

For $x \to 2^{+}$ (that is, for $2 < x < 3$ ), the floor function is strictly equal to the constant integer $⌊ x ⌋ = 2$ .

Therefore, the numerator is exactly zero for all $x \in (2, 3)$ :

\begin{aligned} \lfloor x \rfloor^2 - 4 &= 2^2 - 4 \\ &= 0. \end{aligned}

Since $x^{2} - 4 > 0$ for all $x \in (2, 3)$ , the quotient $\frac{0}{x^{2} - 4}$ is identically equal to $0$ on $(2, 3)$ . We are thus taking the limit of the constant function $0$ :

\begin{aligned} \lim_{x\to 2^+} \frac{\lfloor x \rfloor^2 - 4}{x^2 - 4} &= \lim_{x\to 2^+} 0 \\ &= 0. \end{aligned}

Evaluate the limit or show that it does not exist:

lim_{x \to 2^{-}} \frac{⌊ x ⌋^{2} - 4}{x^{2} - 4}

(Note: $⌊ x ⌋$ is the floor function, which is the same as the greatest integer function $[[x]]$ .)

Answer

$+ \infty$

Solution

For $x \to 2^{-}$ (that is, for $1 < x < 2$ ), the floor function is strictly equal to the constant integer $⌊ x ⌋ = 1$ .

Thus the numerator is constant:

\begin{aligned} \lfloor x \rfloor^2 - 4 &= 1^2 - 4 \\ &= -3. \end{aligned}

As $x \to 2^{-}$ , the denominator $x^{2} - 4 \to 0^{-}$ .

Thus:

\begin{aligned} \lim_{x\to 2^-} \frac{\lfloor x \rfloor^2 - 4}{x^2 - 4} &= \lim_{x\to 2^-} \frac{-3}{x^2 - 4} \\ &= +\infty. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\sqrt{1 + 2 x + x^{2}} - 1}{\sin (2 x)}

Answer

$\frac{1}{2}$

Solution

Direct substitution of $x = 0$ gives $\frac{\sqrt{1} - 1}{\sin (0)} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Notice that $1 + 2 x + x^{2} = (1 + x)^{2}$ , so $\sqrt{1 + 2 x + x^{2}} = | 1 + x |$ .

For x near $0$ (specifically $- 1 < x < 1$ ), $1 + x > 0$ , so $| 1 + x | = 1 + x$ .

The numerator simplifies to $(1 + x) - 1 = x$ :

\begin{aligned} \lim_{x\to 0} \frac{x}{\sin(2x)} &= \lim_{x\to 0} \left( \frac{1}{2} \cdot \frac{2x}{\sin(2x)} \right) \\ &= \frac{1}{2}(1) \\ &= \frac{1}{2}. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\sqrt{1 + 3 x + x^{2}} - 1}{\sin (2 x)}

Answer

$\frac{3}{4}$

Solution

Direct substitution of $x = 0$ gives $\frac{\sqrt{1} - 1}{\sin (0)} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Rationalize the numerator by multiplying numerator and denominator by $\sqrt{1 + 3 x + x^{2}} + 1$ :

\begin{aligned} \frac{(\sqrt{1 + 3x + x^2} - 1)(\sqrt{1 + 3x + x^2} + 1)}{\sin(2x)(\sqrt{1 + 3x + x^2} + 1)} &= \frac{(1 + 3x + x^2) - 1}{\sin(2x)(\sqrt{1 + 3x + x^2} + 1)} \\ &= \frac{3x + x^2}{\sin(2x)(\sqrt{1 + 3x + x^2} + 1)} \\ &= \frac{x(3 + x)}{\sin(2x)(\sqrt{1 + 3x + x^2} + 1)}. \end{aligned}

Dividing numerator and denominator by $x$ for $x \neq 0$ :

\begin{aligned} \lim_{x\to 0} \frac{\sqrt{1 + 3x + x^2} - 1}{\sin(2x)} &= \lim_{x\to 0} \left( \frac{x}{\sin(2x)} \cdot \frac{3 + x}{\sqrt{1 + 3x + x^2} + 1} \right) \\ &= \left( \lim_{x\to 0} \frac{x}{\sin(2x)} \right) \left( \lim_{x\to 0} \frac{3 + x}{\sqrt{1 + 3x + x^2} + 1} \right) \\ &= \left( \frac{1}{2} \right) \cdot \left( \frac{3 + 0}{\sqrt{1 + 0 + 0} + 1} \right) \\ &= \frac{1}{2} \cdot \frac{3}{2} \\ &= \frac{3}{4}. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{\sin (2 x)}

Answer

$\frac{1}{12}$

Solution

Direct substitution of $x = 0$ gives $\frac{\sqrt{9} - 3}{\sin (0)} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Multiply numerator and denominator by $\sqrt{x + 9} + 3$ :

\begin{aligned} \lim_{x\to 0} \frac{\sqrt{x + 9} - 3}{\sin(2x)} &= \lim_{x\to 0} \frac{(\sqrt{x + 9} - 3)(\sqrt{x + 9} + 3)}{\sin(2x)(\sqrt{x + 9} + 3)} \\ &= \lim_{x\to 0} \frac{(x + 9) - 9}{\sin(2x)(\sqrt{x + 9} + 3)} \\ &= \lim_{x\to 0} \left( \frac{x}{\sin(2x)} \cdot \frac{1}{\sqrt{x + 9} + 3} \right) \\ &= \left( \frac{1}{2} \right) \cdot \left( \frac{1}{\sqrt{9} + 3} \right) \\ &= \frac{1}{2} \cdot \frac{1}{6} \\ &= \frac{1}{12}. \end{aligned}

Evaluate the limit:

lim_{t \to 1} \frac{\sqrt{t} - 1}{\sin (1 - t)}

Answer

$- \frac{1}{2}$

Solution

Direct substitution of $t = 1$ gives $\frac{\sqrt{1} - 1}{\sin (0)} = \frac{0}{0}$ , which is an indeterminate form, so direct substitution does not work.

Multiply numerator and denominator by $\sqrt{t} + 1$ :

\begin{aligned} \lim_{t\to 1} \frac{\sqrt{t} - 1}{\sin(1 - t)} &= \lim_{t\to 1} \frac{(\sqrt{t} - 1)(\sqrt{t} + 1)}{\sin(1 - t)(\sqrt{t} + 1)} \\ &= \lim_{t\to 1} \frac{t - 1}{\sin(1 - t)(\sqrt{t} + 1)} \\ &= \lim_{t\to 1} \frac{-(1 - t)}{\sin(1 - t)(\sqrt{t} + 1)}. \end{aligned}

Letting $θ = 1 - t$ , as $t \to 1$ , $θ \to 0$ :

\begin{aligned} \lim_{\theta\to 0} \left( -\frac{\theta}{\sin\theta} \cdot \frac{1}{\sqrt{1 - \theta} + 1} \right) &= (-1) \cdot \frac{1}{1 + 1} \\ &= -\frac{1}{2}. \end{aligned}

Evaluate the limit:

lim_{x \to \infty} \frac{5 x - 3}{2 x + 1}

Answer

$\frac{5}{2}$

Solution

Direct evaluation as $x \to \infty$ gives $\frac{\infty}{\infty}$ , which is an indeterminate form, so direct evaluation does not work.

Using the leading term technique for rational functions as $x \to \infty$ :

\begin{aligned} \lim_{x\to \infty} \frac{5x - 3}{2x + 1} &= \lim_{x\to \infty} \frac{5x}{2x} \\ &= \frac{5}{2}. \end{aligned}

Evaluate the limit:

lim_{x \to \infty} \frac{3 x^{2} - 5 x + 1}{5 x^{2} + x - 9}

Answer

$\frac{3}{5}$

Solution

Direct evaluation as $x \to \infty$ gives $\frac{\infty}{\infty}$ , which is an indeterminate form, so direct evaluation does not work.

Using the leading term technique as $x \to \infty$ :

\begin{aligned} \lim_{x\to \infty} \frac{3x^2 - 5x + 1}{5x^2 + x - 9} &= \lim_{x\to \infty} \frac{3x^2}{5x^2} \\ &= \frac{3}{5}. \end{aligned}

Find:

$lim_{x \to + \infty} \frac{3 + 5 x}{1 - 2 x}$
$lim_{x \to - \infty} \frac{3 x^{3} - 4 x^{2} + x + 3}{- x^{3} + 2 x^{2}}$
$lim_{x \to + \infty} \frac{15 + 4 x - 5 x^{3} + 3 x^{4}}{1 + 16 x^{2} - x^{5}}$
$lim_{x \to - \infty} \frac{x^{7} + 1}{- x^{6} + x^{3} + 9}$

Answer

(a) $- \frac{5}{2}$ (b) $- 3$ (c) $0$ (d) $+ \infty$

Solution

In each part, direct evaluation as $x \to \pm \infty$ yields an indeterminate form ( $\frac{\infty}{\infty}$ or $\infty - \infty$ ), so direct evaluation does not work. We keep only the largest-degree terms in numerator and denominator.

(a)

\begin{aligned} \lim_{x\to +\infty} \frac{3+5x}{1-2x} &= \lim_{x\to +\infty} \frac{5x}{-2x} \\ &= -\frac{5}{2}. \end{aligned}

(b)

\begin{aligned} \lim_{x\to -\infty} \frac{3x^3 - 4x^2 + x + 3}{-x^3 + 2x^2} &= \lim_{x\to -\infty} \frac{3x^3}{-x^3} \\ &= \lim_{x\to -\infty} (-3) \\ &= -3. \end{aligned}

(c)

\begin{aligned} \lim_{x\to +\infty} \frac{15 + 4x - 5x^3 + 3x^4}{1 + 16x^2 - x^5} &= \lim_{x\to +\infty} \frac{3x^4}{-x^5} \\ &= \lim_{x\to +\infty} \frac{-3}{x} \\ &= -3 \lim_{x\to +\infty} \frac{1}{x} \\ &= -3(0) \\ &= 0. \end{aligned}

(d)

\begin{aligned} \lim_{x\to -\infty} \frac{x^7 + 1}{-x^6 + x^3 + 9} &= \lim_{x\to -\infty} \frac{x^7}{-x^6} \\ &= \lim_{x\to -\infty} (-x) \\ &= -\lim_{x\to -\infty} x \\ &= -(-\infty) \\ &= +\infty. \end{aligned}

Evaluate the limit:

lim_{x \to \infty} \frac{8 x^{3} - 1}{3 x + 1}

Answer

$+ \infty$

Solution

Direct evaluation as $x \to \infty$ gives $\frac{\infty}{\infty}$ , which is an indeterminate form, so direct evaluation does not work.

Using the leading term technique:

\begin{aligned} \lim_{x\to \infty} \frac{8x^3 - 1}{3x + 1} &= \lim_{x\to \infty} \frac{8x^3}{3x} \\ &= \lim_{x\to \infty} \left( \frac{8}{3} x^2 \right) \\ &= +\infty. \end{aligned}

Evaluate the limit:

lim_{x \to - \infty} \frac{2 x^{4} + 3 x^{3} - 10 x + 1}{10 x^{3} + 11 x^{2} + 9}

Answer

$- \infty$

Solution

Direct evaluation as $x \to - \infty$ gives $\frac{\infty}{- \infty}$ , which is an indeterminate form, so direct evaluation does not work.

Using the leading term technique as $x \to - \infty$ :

\begin{aligned} \lim_{x\to -\infty} \frac{2x^4 + 3x^3 - 10x + 1}{10x^3 + 11x^2 + 9} &= \lim_{x\to -\infty} \frac{2x^4}{10x^3} \\ &= \lim_{x\to -\infty} \left( \frac{1}{5} x \right) \\ &= -\infty. \end{aligned}

Find $lim_{x \to + \infty} (12 + 9 x + 3 x^{3} + 1200 x^{5} - x^{6})$ .

Answer

$- \infty$

Solution

Direct evaluation as $x \to + \infty$ gives $\infty - \infty$ , which is an indeterminate form, so direct evaluation does not work.

Using the leading term technique:

\begin{aligned} \lim_{x\to +\infty} (12 + 9x + 3x^3 + 1200x^5 - x^6) &= \lim_{x\to +\infty} (-x^6) \\ &= -\lim_{x\to +\infty} x^6 \\ &= -(+\infty) \\ &= -\infty. \end{aligned}

Evaluate the limit:

lim_{x \to - \infty} \frac{- x + 1}{x^{2} + 3 x - 7} sgn (x)

Answer

$0$

Solution

As $x \to - \infty$ , $x < 0$ , so $sgn (x) = - 1$ .

Direct evaluation of $\frac{- x + 1}{x^{2} + 3 x - 7}$ gives $\frac{\infty}{\infty}$ , which is an indeterminate form, so direct evaluation does not work.

Thus:

\begin{aligned} \lim_{x\to -\infty} \frac{-x + 1}{x^2 + 3x - 7}(-1) &= \lim_{x\to -\infty} \frac{x - 1}{x^2 + 3x - 7} \\ &= \lim_{x\to -\infty} \frac{x}{x^2} \\ &= \lim_{x\to -\infty} \frac{1}{x} \\ &= 0. \end{aligned}

Evaluate the limit:

lim_{x \to \infty} [(\frac{1}{x} - 1) (\frac{4 x^{2} - 1}{3 x^{2} + 1})]

Answer

$- \frac{4}{3}$

Solution

Direct evaluation of the second factor gives $\frac{\infty}{\infty}$ , which is an indeterminate form, so we evaluate the limit of each factor using leading terms:

First factor: $lim_{x \to \infty} (\frac{1}{x} - 1) = 0 - 1 = - 1$ .
Second factor: $lim_{x \to \infty} \frac{4 x^{2} - 1}{3 x^{2} + 1} = lim_{x \to \infty} \frac{4 x^{2}}{3 x^{2}} = \frac{4}{3}$ .

Multiplying the limits:\begin{aligned} (-1) \cdot \frac{4}{3} &= -\frac{4}{3}. \end{aligned}

Evaluate the limit:

lim_{x \to \infty} \frac{\sqrt{x - 1}}{\sqrt{4 x + 3}}

Answer

$\frac{1}{2}$

Solution

Direct evaluation gives $\frac{\infty}{\infty}$ , which is an indeterminate form, so direct evaluation does not work.

Combine the square roots under a single radical:

\begin{aligned} \lim_{x\to \infty} \frac{\sqrt{x - 1}}{\sqrt{4x + 3}} &= \lim_{x\to \infty} \sqrt{\frac{x - 1}{4x + 3}} \\ &= \sqrt{\lim_{x\to \infty} \frac{x - 1}{4x + 3}} \\ &= \sqrt{\frac{1}{4}} \\ &= \frac{1}{2}. \end{aligned}

Find $lim_{x \to + \infty} \frac{x^{2} - 4 \sqrt{x} + 3 x}{- 3 x^{2} + 5 x^{2 / 3} + 1}$ .

Answer

$- \frac{1}{3}$

Solution

Direct evaluation gives $\frac{\infty}{- \infty}$ , which is an indeterminate form, so direct evaluation does not work.

Factor out $x^{2}$ :

\begin{aligned} \lim_{x\to +\infty} \frac{x^2 - 4\sqrt{x} + 3x}{-3x^2 + 5x^{2/3} + 1} &= \lim_{x\to +\infty} \frac{x^2 \left( 1 - \frac{4}{x^{3/2}} + \frac{3}{x} \right)}{x^2 \left( -3 + \frac{5}{x^{4/3}} + \frac{1}{x^2} \right)} \\ &= \lim_{x\to +\infty} \frac{1 - \frac{4}{x^{3/2}} + \frac{3}{x}}{-3 + \frac{5}{x^{4/3}} + \frac{1}{x^2}} \\ &= \frac{1 - 4(0) + 3(0)}{-3 + 5(0) + 0} \\ &= -\frac{1}{3}. \end{aligned}

Find $lim_{x \to - \infty} \frac{\sqrt{4 x^{4} + 3 x^{2} + 1}}{1 - 2 x^{2}}$ .

Answer

$- 1$

Solution

Direct evaluation as $x \to - \infty$ gives $\frac{\infty}{- \infty}$ , which is an indeterminate form, so direct evaluation does not work.

Factor out $x^{4}$ inside radical and $x^{2}$ in denominator:

\begin{aligned} \lim_{x\to -\infty} \frac{\sqrt{4x^4 + 3x^2 + 1}}{1 - 2x^2} &= \lim_{x\to -\infty} \frac{\sqrt{x^4 \left( 4 + \frac{3}{x^2} + \frac{1}{x^4} \right)}}{x^2 \left( \frac{1}{x^2} - 2 \right)} \\ &= \lim_{x\to -\infty} \frac{x^2 \sqrt{4 + \frac{3}{x^2} + \frac{1}{x^4}}}{x^2 \left( \frac{1}{x^2} - 2 \right)} \\ &= \lim_{x\to -\infty} \frac{\sqrt{4 + \frac{3}{x^2} + \frac{1}{x^4}}}{\frac{1}{x^2} - 2} \\ &= \frac{\sqrt{4 + 0 + 0}}{0 - 2} \\ &= \frac{2}{-2} \\ &= -1. \end{aligned}

Graph of f(x) = sqrt(4x^4+3x^2+1)/(1-2x^2) — Graph of $f (x) = \frac{\sqrt{4 x^{4} + 3 x^{2} + 1}}{1 - 2 x^{2}}$ . The limit as $x \to \pm \infty$ is $- 1$ .

Find:

$lim_{t \to + \infty} \frac{\sqrt{t^{2} + 6}}{3 t - 9}$
$lim_{t \to - \infty} \frac{\sqrt{t^{2} + 6}}{3 t - 9}$

Answer

(a)

\frac{1}{3}

(b)

- \frac{1}{3}

Solution

Direct evaluation yields

\frac{\infty}{\infty}

\frac{\infty}{- \infty}

, which are indeterminate forms, so direct evaluation does not work.

(a) For

t > 0

\sqrt{t^{2}} = | t | = t

: \begin{aligned} \lim_{t\to +\infty} \frac{\sqrt{t^2 + 6}}{3t - 9} &= \lim_{t\to +\infty} \frac{t\sqrt{1 + \frac{6}{t^2}}}{t\left(3 - \frac{9}{t}\right)} \\ &= \frac{\sqrt{1 + 0}}{3 - 0} \\ &= \frac{1}{3}. \end{aligned} (b) For

t < 0

\sqrt{t^{2}} = | t | = - t

: \begin{aligned} \lim_{t\to -\infty} \frac{\sqrt{t^2 + 6}}{3t - 9} &= \lim_{t\to -\infty} \frac{-t\sqrt{1 + \frac{6}{t^2}}}{t\left(3 - \frac{9}{t}\right)} \\ &= \frac{-\sqrt{1 + 0}}{3 - 0} \\ &= -\frac{1}{3}. \end{aligned}

Graph of f(t) = sqrt(t^2+6)/(3t-9) — Graph of $f (t) = \frac{\sqrt{t^{2} + 6}}{3 t - 9}$ . Limit as $t \to + \infty$ is $1 / 3$ and as $t \to - \infty$ is $- 1 / 3$ .

Evaluate the limit:

lim_{x \to \infty} (\frac{1}{⌊ x ⌋} - \frac{2}{\sqrt{x}})

Answer

$0$

Solution

As $x \to \infty$ , $x - 1 < ⌊ x ⌋ \leq x ⟹ ⌊ x ⌋ \to \infty$ .

Thus:

\begin{aligned} \lim_{x\to \infty} \frac{1}{\lfloor x \rfloor} &= 0 \\ \lim_{x\to \infty} \frac{2}{\sqrt{x}} &= 0. \end{aligned}

Subtracting the limits:

\begin{aligned} \lim_{x\to \infty} \left( \frac{1}{\lfloor x \rfloor} - \frac{2}{\sqrt{x}} \right) &= 0 - 0 \\ &= 0. \end{aligned}

Since this is form $0 - 0 = 0$ , it is not an indeterminate form.

Find $lim_{x \to + \infty} \frac{3 x - 4}{5 x - \sin (x^{2})}$ .

Answer

$\frac{3}{5}$

Solution

Direct evaluation yields $\frac{\infty}{\infty}$ , which is an indeterminate form, so direct evaluation does not work.

Since $- 1 \leq - \sin (x^{2}) \leq 1$ , for $x$ large enough that $3 x - 4 > 0$ :

\frac{3 x - 4}{5 x + 1} \leq \frac{3 x - 4}{5 x - \sin (x^{2})} \leq \frac{3 x - 4}{5 x - 1} .

Evaluating the outer limits as $x \to + \infty$ :

\begin{aligned} \lim_{x\to +\infty} \frac{3x - 4}{5x - 1} &= \frac{3}{5} \\ \lim_{x\to +\infty} \frac{3x - 4}{5x + 1} &= \frac{3}{5}. \end{aligned}

By the Sandwich Theorem:

\begin{aligned} \lim_{x\to +\infty} \frac{3x - 4}{5x - \sin(x^2)} &= \frac{3}{5}. \end{aligned}

Find $lim_{x \to + \infty} (\frac{2 x^{3}}{x^{2} - 4} - \frac{2 x^{2}}{x - 1})$ .

Answer

$- 2$

Solution

Using a dominant term argument as $x \to + \infty$ :

\begin{aligned} \lim_{x\to +\infty} \frac{2x^3}{x^2 - 4} &= \lim_{x\to +\infty} \frac{2x^3}{x^2} \\ &= \lim_{x\to +\infty} 2x \\ &= +\infty \end{aligned}

and

\begin{aligned} \lim_{x\to +\infty} \frac{2x^2}{x - 1} &= \lim_{x\to +\infty} \frac{2x^2}{x} \\ &= \lim_{x\to +\infty} 2x \\ &= +\infty. \end{aligned}

Thus, evaluating the terms separately gives the indeterminate form $\infty - \infty$ , so we cannot use the Difference Law to evaluate the limits separately. We must combine the two fractions into a single rational function:

\begin{aligned} \frac{2x^3}{x^2 - 4} - \frac{2x^2}{x - 1} &= \frac{2x^3(x - 1) - 2x^2(x^2 - 4)}{(x^2 - 4)(x - 1)} \\ &= \frac{2x^4 - 2x^3 - 2x^4 + 8x^2}{x^3 - x^2 - 4x + 4} \\ &= \frac{-2x^3 + 8x^2}{x^3 - x^2 - 4x + 4}. \end{aligned}

Now taking the limit as $x \to + \infty$ using the leading term technique:

\begin{aligned} \lim_{x\to +\infty} \left( \frac{2x^3}{x^2 - 4} - \frac{2x^2}{x - 1} \right) &= \lim_{x\to +\infty} \frac{-2x^3 + 8x^2}{x^3 - x^2 - 4x + 4} \\ &= \lim_{x\to +\infty} \frac{-2x^3}{x^3} \\ &= -2. \end{aligned}

Evaluate the limit:

lim_{x \to \infty} (\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1})

Answer

$0$

Solution

As $x \to \infty$ , $\sqrt{x^{2} + 1} \to + \infty$ and $\sqrt{x^{2} - 1} \to + \infty$ . Subtracting these limits yields the indeterminate form $\infty - \infty$ , so direct evaluation does not work.

To resolve this indeterminate form, multiply and divide by the conjugate radical $\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1}$ :

\begin{aligned} \lim_{x\to \infty} \left( \sqrt{x^2 + 1} - \sqrt{x^2 - 1} \right) &= \lim_{x\to \infty} \frac{(\sqrt{x^2 + 1} - \sqrt{x^2 - 1})(\sqrt{x^2 + 1} + \sqrt{x^2 - 1})}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}} \\ &= \lim_{x\to \infty} \frac{(x^2 + 1) - (x^2 - 1)}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}} \\ &= \lim_{x\to \infty} \frac{2}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}. \end{aligned}

As $x \to \infty$ , the numerator remains $2$ while the denominator approaches $+ \infty$ , so the limit is:

\begin{aligned} \lim_{x\to \infty} \frac{2}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}} &= 0. \end{aligned}

Find:

$lim_{x \to - \infty} (\sqrt{x^{4} + 5} - 2 x^{2})$
$lim_{x \to - \infty} (\sqrt{x^{4} + 5 x^{2}} - x^{2})$
$lim_{x \to - \infty} (\sqrt{2 x^{2} - 1} - 7 x)$

Answer

(a)

- \infty

(b)

\frac{5}{2}

(c)

+ \infty

Solution

(a) Direct evaluation as

x \to - \infty

yields

\infty - \infty

, which is an indeterminate form, so direct evaluation does not work. Factoring out

x^{2}

: \begin{aligned} \lim_{x\to -\infty} \left( \sqrt{x^4 + 5} - 2x^2 \right) &= \lim_{x\to -\infty} \left[ x^2 \left( \sqrt{1 + \frac{5}{x^4}} - 2 \right) \right] \\ &= \left( \lim_{x\to -\infty} x^2 \right) \left( \sqrt{\lim_{x\to -\infty} \left( 1 + \frac{5}{x^4} \right)} - 2 \right) \\ &= (+\infty)(\sqrt{1 + 0} - 2) \\ &= (+\infty)(-1) \\ &= -\infty. \end{aligned}

Graph of f(x) = sqrt(x^4+5)-2x^2 — Graph of $f (x) = \sqrt{x^{4} + 5} - 2 x^{2}$ . Limit as $x \to \pm \infty$ is $- \infty$ .

(b) Direct evaluation as

x \to - \infty

yields

\infty - \infty

, which is an indeterminate form, so direct evaluation does not work. Multiply and divide by conjugate: \begin{aligned} \lim_{x\to -\infty} \left( \sqrt{x^4 + 5x^2} - x^2 \right) &= \lim_{x\to -\infty} \frac{(\sqrt{x^4 + 5x^2} - x^2)(\sqrt{x^4 + 5x^2} + x^2)}{\sqrt{x^4 + 5x^2} + x^2} \\ &= \lim_{x\to -\infty} \frac{x^4 + 5x^2 - x^4}{\sqrt{x^4 + 5x^2} + x^2} \\ &= \lim_{x\to -\infty} \frac{5x^2}{x^2 \left( \sqrt{1 + \frac{5}{x^2}} + 1 \right)} \\ &= \frac{5}{\sqrt{1 + 0} + 1} \\ &= \frac{5}{2}. \end{aligned}

Graph of f(x) = sqrt(x^4+5x^2)-x^2 — Graph of $f (x) = \sqrt{x^{4} + 5 x^{2}} - x^{2}$ . Limit as $x \to \pm \infty$ is $2.5$ .

(c) Because

x < 0

\sqrt{x^{2}} = | x | = - x

: \begin{aligned} \lim_{x\to -\infty} \sqrt{2x^2 - 1} &= \lim_{x\to -\infty} \left( -x \sqrt{2 - \frac{1}{x^2}} \right) \\ &= -(-\infty)\sqrt{2 - 0} \\ &= +\infty, \end{aligned} and

lim_{x \to - \infty} (- 7 x) = + \infty

. Thus: \begin{aligned} \lim_{x\to -\infty} \left( \sqrt{2x^2 - 1} - 7x \right) &= (+\infty) + (+\infty) \\ &= +\infty. \end{aligned} Since this is form

(+ \infty) + (+ \infty) = + \infty

, it is not an indeterminate form.

Graph of y = sqrt(2x^2-1)-7x — From this graph, it is clear that $y = \sqrt{2 x^{2} - 1} - 7 x \to + \infty$ as $x \to - \infty$ .

Quick Reference

Exercises

Quick Links

Social Media