The Number E And Some Certain Limits

In this section, we examine the fundamental limit definition of Euler's number e ( $\approx 2.71828$ ) and derive essential logarithmic, exponential, and power function limits that form the backbone of differential calculus.

Quick Reference

Limit Formula	Value	Substitution / Method
Definition of e at $0$	$lim_{x \to 0} (1 + x)^{1 / x} = e$	Base definition of Euler's constant
Definition of e at $\infty$	$lim_{x \to \pm \infty} {(1 + \frac{1}{x})}^{x} = e$	Variable change $u = \frac{1}{x}$
Logarithmic Limit at $0$	$lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1$	Rewrite $\frac{1}{x} \ln (1 + x) = \ln [(1 + x)^{1 / x}]$
Exponential Limit at $0$	$lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$	Substitution $u = e^{x} - 1 ⟹ x = \ln (1 + u)$
Power Derivative Base Limit	$lim_{x \to 0} \frac{(1 + x)^{r} - 1}{x} = r$	Substitution $u = (1 + x)^{r} - 1$

The Number e

One of the most important limits in calculus is

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\lim_{x\to 0} (1 + x)^{1/x} = e}

where e is an irrational number (like $π$ ) and is approximately $2.718281828$ .

We are not going to prove that such a limit exists, but we will content ourselves by plotting $y = (1 + x)^{1 / x}$ (the following figure), and show graphically that as $x \approx 0$ , the function $y = (1 + x)^{1 / x}$ takes on values in the near neighborhood of $2.718 \dots$ , and therefore $e \approx 2.7182$ .

Graph of y = (1+x)^(1/x). Note that this function is not defined at x = 0 — Graph of $y = (1 + x)^{1 / x}$ . Note that this function is not defined at $x = 0$ .

As x approaches zero from the left, $y$ decreases and approaches e as a limit. As x approaches zero from the right, $y$ increases and also approaches e as a limit.

As $x \to \infty$ , $y$ approaches the limit 1, and as $x \to - 1$ from the right, $y$ increases indefinitely (see the following table).

x	y	x	y
$100$	$1.04723$
$10$	$1.27098$	$- 0.99$	$104.762$
$1$	$2.0000$	$- 0.9$	$12.9155$
$0.5$	$2.2500$	$- 0.5$	$4.0000$
$0.1$	$2.59374$	$- 0.1$	$2.86797$
$0.01$	$2.70481$	$- 0.01$	$2.732$
$0.001$	$2.71692$	$- 0.001$	$2.71964$
$0.0001$	$2.71815$	$- 0.0001$	$2.71842$

Let's put $u = 1 / x$ . As x approaches 0 from the positive side ( $x \to 0^{+}$ ), u approaches positive infinity ( $u \to + \infty$ ). Therefore,

lim_{x \to 0^{+}} (1 + x)^{1 / x} = lim_{u \to + \infty} {(1 + \frac{1}{u})}^{u} = e .

With a change in notation, we can write this result as follows:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\lim_{x\to +\infty} \left(1 + \frac{1}{x}\right)^x = e}

What about the limit of

{(1 + \frac{1}{x})}^{x}

x \to - \infty

Similarly as $x \to 0^{-}$ , $u \to - \infty$ . Therefore,

lim_{x \to 0^{-}} (1 + x)^{1 / x} = lim_{u \to - \infty} {(1 + \frac{1}{u})}^{u} = e,

lim_{x \to - \infty} {(1 + \frac{1}{x})}^{x} = e .

Graph of y = (1 + 1/x)^x. As x - loading= — Graph of $y = {(1 + \frac{1}{x})}^{x}$ . As $x \to \pm \infty$ , $y \to e \approx 2.71828$ .

Computation of Certain Limits

Now we know that $lim_{x \to 0} (1 + x)^{1 / x} = e$ , by applying the continuity of some basic functions, we can determine a number of significant limits that will be necessary for the next chapter.

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\lim_{x\to 0} \frac{\ln(1 + x)}{x} = 1}\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\lim_{x\to 0} \frac{e^x - 1}{x} = 1}\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\lim_{x\to 0} \frac{(1 + x)^r - 1}{x} = r}

We have:

\frac{1}{x} \ln (1 + x) = \ln [(1 + x)^{\frac{1}{x}}]

It follows from the continuity of the logarithmic function that:

\begin{aligned} \lim_{x \to 0} \ln \left[(1 + x)^{\frac{1}{x}}\right] &= \ln \left[\lim_{x \to 0} (1 + x)^{\frac{1}{x}}\right] \\ &= \ln e \\ &= 1. \end{aligned}

For the second limit, let $e^{x} - 1 = u$ or:

e^{x} = u + 1.

Taking the natural log of both sides provides:

x = \ln (u + 1) .

Therefore, we can rewrite the original expression as:

\frac{e^{x} - 1}{x} = \frac{u}{\ln (u + 1)} .

As $x \to 0$ , $u = e^{x} - 1 \to 0$ , leading to:

\begin{aligned} \lim_{x \to 0} \frac{e^x - 1}{x} &= \lim_{u \to 0} \frac{u}{\ln (1 + u)} \\ &= \frac{1}{\displaystyle \lim_{u \to 0} \frac{\ln (1 + u)}{u}} \\ &= \frac{1}{1} \\ &= 1. \end{aligned}

To prove $lim_{x \to 0} \frac{(1 + x)^{r} - 1}{x} = r$ , let:

(1 + x)^{r} - 1 = u

or:

(1 + x)^{r} = 1 + u .

Taking logarithms of both sides, we obtain:

r \ln (1 + x) = \ln (1 + u) .

Therefore:

\begin{aligned} \frac{(1 + x)^r - 1}{x} &= \frac{u}{x} \\ &= \frac{u}{x} \times \frac{r \ln (1 + x)}{\ln (1 + u)} \\ &= r \frac{u}{\ln (1 + u)} \frac{\ln (1 + x)}{x}. \end{aligned}

It follows from the continuity of a power function that as $x \to 0$ :

u = (1 + x)^{r} - 1 \to 0.

Therefore, we conclude:

\begin{aligned} \lim_{x \to 0} \frac{(1 + x)^r - 1}{x} &= r \lim_{u \to 0} \frac{u}{\ln (1 + u)} \cdot \lim_{x \to 0} \frac{\ln (1 + x)}{x} \\ &= r \times 1 \times 1 \\ &= r. \end{aligned}

Exercises

Evaluate the limit:

lim_{x \to 0} (1 + 3 x)^{1 / x}

Answer

$e^{3}$

Solution

Let $u = 3 x$ . As $x \to 0$ , we have $u \to 0$ , and $\frac{1}{x} = \frac{3}{u}$ .

Substituting $u$ into the limit:

\begin{aligned} \lim_{x\to 0} (1 + 3x)^{1/x} &= \lim_{u\to 0} (1 + u)^{3/u} \\ &= \lim_{u\to 0} \left[ (1 + u)^{1/u} \right]^3. \end{aligned}

Since $lim_{u \to 0} (1 + u)^{1 / u} = e$ , by the power rule for limits:

\begin{aligned} \lim_{x\to 0} (1 + 3x)^{1/x} &= \left[ \lim_{u\to 0} (1 + u)^{1/u} \right]^3 \\ &= e^3. \end{aligned}

Evaluate the limit at infinity:

lim_{x \to + \infty} {(1 + \frac{5}{x})}^{2 x}

Answer

$e^{10}$

Solution

Let $u = \frac{x}{5}$ . As $x \to + \infty$ , we have $u \to + \infty$ , and $x = 5 u$ .

Substituting $u$ into the limit:

\begin{aligned} \lim_{x\to +\infty} \left(1 + \frac{5}{x}\right)^{2x} &= \lim_{u\to +\infty} \left(1 + \frac{1}{u}\right)^{2(5u)} \\ &= \lim_{u\to +\infty} \left[ \left(1 + \frac{1}{u}\right)^u \right]^{10}. \end{aligned}

Since $lim_{u \to + \infty} {(1 + \frac{1}{u})}^{u} = e$ , we obtain:

\begin{aligned} \lim_{x\to +\infty} \left(1 + \frac{5}{x}\right)^{2x} &= \left[ \lim_{u\to +\infty} \left(1 + \frac{1}{u}\right)^u \right]^{10} \\ &= e^{10}. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{e^{4 x} - 1}{3 x}

Answer

$\frac{4}{3}$

Solution

Multiply and divide by $4$ :

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

Let $u = 4 x$ . As $x \to 0$ , $u \to 0$ . Using the standard exponential limit $lim_{u \to 0} \frac{e^{u} - 1}{u} = 1$ :

\begin{aligned} \lim_{x\to 0} \frac{e^{4x} - 1}{3x} &= \frac{4}{3} \lim_{u\to 0} \frac{e^u - 1}{u} \\ &= \frac{4}{3}(1) \\ &= \frac{4}{3}. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{2^{x} - 1}{x}

Answer

$\ln 2$

Solution

Recall the identity $2^{x} = e^{x \ln 2}$ .

Substituting this into the limit yields:

lim_{x \to 0} \frac{2^{x} - 1}{x} = lim_{x \to 0} \frac{e^{x \ln 2} - 1}{x} .

Let $u = x \ln 2$ . As $x \to 0$ , $u \to 0$ , and $x = \frac{u}{\ln 2}$ .

Therefore:

\begin{aligned} \lim_{x\to 0} \frac{2^x - 1}{x} &= \lim_{u\to 0} \frac{e^u - 1}{\frac{u}{\ln 2}} \\ &= \ln 2 \cdot \lim_{u\to 0} \frac{e^u - 1}{u} \\ &= (\ln 2)(1) \\ &= \ln 2. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\ln (1 + 5 x)}{2 x}

Answer

$\frac{5}{2}$

Solution

Multiply and divide by $5$ :

\begin{aligned} \lim_{x\to 0} \frac{\ln(1 + 5x)}{2x} &= \lim_{x\to 0} \left( \frac{5}{2} \cdot \frac{\ln(1 + 5x)}{5x} \right) \\ &= \frac{5}{2} \lim_{x\to 0} \frac{\ln(1 + 5x)}{5x}. \end{aligned}

Let $u = 5 x$ . As $x \to 0$ , $u \to 0$ . Using the standard logarithmic limit $lim_{u \to 0} \frac{\ln (1 + u)}{u} = 1$ :

\begin{aligned} \lim_{x\to 0} \frac{\ln(1 + 5x)}{2x} &= \frac{5}{2} \lim_{u\to 0} \frac{\ln(1 + u)}{u} \\ &= \frac{5}{2}(1) \\ &= \frac{5}{2}. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\sqrt{1 + 3 x} - 1}{x}

Answer

$\frac{3}{2}$

Solution

Rewrite $\sqrt{1 + 3 x} = (1 + 3 x)^{1 / 2}$ .

Using the power limit rule $lim_{t \to 0} \frac{(1 + t)^{r} - 1}{t} = r$ with $r = 1 / 2$ :

\begin{aligned} \lim_{x\to 0} \frac{(1 + 3x)^{1/2} - 1}{x} &= \lim_{x\to 0} \left( 3 \cdot \frac{(1 + 3x)^{1/2} - 1}{3x} \right) \\ &= 3 \lim_{u\to 0} \frac{(1 + u)^{1/2} - 1}{u} \qquad (\text{where } u = 3x) \\ &= 3 \left( \frac{1}{2} \right) \\ &= \frac{3}{2}. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\sin (5 x)}{\ln (1 + 3 x)}

Answer

$\frac{5}{3}$

Solution

Divide both the numerator and denominator by $x$ (for $x \neq 0$ ):

lim_{x \to 0} \frac{\sin (5 x)}{\ln (1 + 3 x)} = lim_{x \to 0} \frac{\frac{\sin (5 x)}{x}}{\frac{\ln (1 + 3 x)}{x}} .

Evaluating the limit of the numerator:

\begin{aligned} \lim_{x\to 0} \frac{\sin(5x)}{x} &= \lim_{x\to 0} \left( 5 \cdot \frac{\sin(5x)}{5x} \right) \\ &= 5(1) = 5. \end{aligned}

Evaluating the limit of the denominator:

\begin{aligned} \lim_{x\to 0} \frac{\ln(1 + 3x)}{x} &= \lim_{x\to 0} \left( 3 \cdot \frac{\ln(1 + 3x)}{3x} \right) \\ &= 3(1) = 3. \end{aligned}

Therefore:

\begin{aligned} \lim_{x\to 0} \frac{\sin(5x)}{\ln(1 + 3x)} &= \frac{5}{3}. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{e^{2 x} - 1}{\tan (4 x)}

Answer

$\frac{1}{2}$

Solution

Divide both the numerator and denominator by $x$ (for $x \neq 0$ ):

lim_{x \to 0} \frac{e^{2 x} - 1}{\tan (4 x)} = lim_{x \to 0} \frac{\frac{e^{2 x} - 1}{x}}{\frac{\tan (4 x)}{x}} .

Evaluating the numerator limit:

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

Evaluating the denominator limit:

\begin{aligned} \lim_{x\to 0} \frac{\tan(4x)}{x} &= 4 \lim_{x\to 0} \frac{\tan(4x)}{4x} \\ &= 4(1) = 4. \end{aligned}

Therefore:

\begin{aligned} \lim_{x\to 0} \frac{e^{2x} - 1}{\tan(4x)} &= \frac{2}{4} \\ &= \frac{1}{2}. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{e^{3 x} - e^{- 2 x}}{x}

Answer

$5$

Solution

Add and subtract $1$ in the numerator:

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

Split the quotient into two fractions:

\frac{e^{3 x} - e^{- 2 x}}{x} = \frac{e^{3 x} - 1}{x} - \frac{e^{- 2 x} - 1}{x} .

Taking the limits individually:

\begin{aligned} \lim_{x\to 0} \frac{e^{3x} - 1}{x} &= 3 \lim_{x\to 0} \frac{e^{3x} - 1}{3x} \\ &= 3(1) = 3, \end{aligned}

and:

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

Subtracting the two limits:

\begin{aligned} \lim_{x\to 0} \frac{e^{3x} - e^{-2x}}{x} &= 3 - (-2) \\ &= 5. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\ln (1 + \sin (2 x))}{x}

Answer

$2$

Solution

Multiply and divide by $\sin (2 x)$ :

\frac{\ln (1 + \sin (2 x))}{x} = \frac{\ln (1 + \sin (2 x))}{\sin (2 x)} \cdot \frac{\sin (2 x)}{x} .

Let $u = \sin (2 x)$ . As $x \to 0$ , $u \to 0$ .

Taking the limit of each factor:

\begin{aligned} \lim_{x\to 0} \frac{\ln(1 + \sin(2x))}{\sin(2x)} &= \lim_{u\to 0} \frac{\ln(1 + u)}{u} \\ &= 1, \end{aligned}

and:

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

Therefore:

\begin{aligned} \lim_{x\to 0} \frac{\ln(1 + \sin(2x))}{x} &= (1)(2) \\ &= 2. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\ln (1 + 2 x)}{\sin (3 x)}

Answer

$\frac{2}{3}$

Solution

Divide numerator and denominator by $x$ (for $x \neq 0$ ):

lim_{x \to 0} \frac{\ln (1 + 2 x)}{\sin (3 x)} = lim_{x \to 0} \frac{\frac{\ln (1 + 2 x)}{x}}{\frac{\sin (3 x)}{x}} .

Evaluating the numerator limit:

\begin{aligned} \lim_{x\to 0} \frac{\ln(1 + 2x)}{x} &= 2 \lim_{x\to 0} \frac{\ln(1 + 2x)}{2x} \\ &= 2(1) = 2. \end{aligned}

Evaluating the denominator limit:

\begin{aligned} \lim_{x\to 0} \frac{\sin(3x)}{x} &= 3 \lim_{x\to 0} \frac{\sin(3x)}{3x} \\ &= 3(1) = 3. \end{aligned}

Therefore:

\begin{aligned} \lim_{x\to 0} \frac{\ln(1 + 2x)}{\sin(3x)} &= \frac{2}{3}. \end{aligned}

Evaluate the limit at infinity:

lim_{x \to + \infty} {(\frac{x + 4}{x - 1})}^{x}

Answer

$e^{5}$

Solution

Rewrite the fraction $\frac{x + 4}{x - 1}$ by performing long division:

\frac{x + 4}{x - 1} = \frac{(x - 1) + 5}{x - 1} = 1 + \frac{5}{x - 1} .

Substituting this into the limit expression:

{(\frac{x + 4}{x - 1})}^{x} = {(1 + \frac{5}{x - 1})}^{x} .

Express the exponent $x$ in terms of $(x - 1)$ :

{(1 + \frac{5}{x - 1})}^{x} = {(1 + \frac{5}{x - 1})}^{x - 1} \cdot (1 + \frac{5}{x - 1}) .

Let $u = x - 1$ . As $x \to + \infty$ , $u \to + \infty$ .

Evaluating the limits:

\begin{aligned} \lim_{u\to +\infty} \left(1 + \frac{5}{u}\right)^u &= e^5, \end{aligned}

and:

\begin{aligned} \lim_{u\to +\infty} \left(1 + \frac{5}{u}\right) &= 1 + 0 = 1. \end{aligned}

Therefore:

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

Evaluate the limit:

lim_{x \to 0} (1 + \tan x)^{\csc x}

Answer

$e$

Solution

Recall that $\csc x = \frac{1}{\sin x}$ .

Rewrite the expression:

(1 + \tan x)^{\csc x} = {[(1 + \tan x)^{\frac{1}{\tan x}}]}^{\frac{\tan x}{\sin x}} .

Simplify the exponent $\frac{\tan x}{\sin x}$ :

\frac{\tan x}{\sin x} = \frac{\frac{\sin x}{\cos x}}{\sin x} = \frac{1}{\cos x} .

Let $u = \tan x$ . As $x \to 0$ , $u \to 0$ .

Taking the limit of the base:

\begin{aligned} \lim_{x\to 0} (1 + \tan x)^{\frac{1}{\tan x}} &= \lim_{u\to 0} (1 + u)^{1/u} \\ &= e. \end{aligned}

Taking the limit of the exponent:

\begin{aligned} \lim_{x\to 0} \frac{1}{\cos x} &= \frac{1}{1} = 1. \end{aligned}

Combining the base and exponent limits:

\begin{aligned} \lim_{x\to 0} (1 + \tan x)^{\csc x} &= e^1 \\ &= e. \end{aligned}

Evaluate the limit:

lim_{x \to 0} (\cos x)^{1 / x^{2}}

Answer

$e^{- 1 / 2}$

Solution

Rewrite $\cos x = 1 + (\cos x - 1)$ .

Substituting into the limit expression:

(\cos x)^{1 / x^{2}} = {[1 + (\cos x - 1)]}^{1 / x^{2}} .

Multiply and divide the exponent by $(\cos x - 1)$ :

{[1 + (\cos x - 1)]}^{1 / x^{2}} = {[{(1 + (\cos x - 1))}^{\frac{1}{\cos x - 1}}]}^{\frac{\cos x - 1}{x^{2}}} .

Let $u = \cos x - 1$ . As $x \to 0$ , $u \to 0$ .

Taking the limit of the inner base:

\begin{aligned} \lim_{x\to 0} \left( 1 + u \right)^{1/u} &= e. \end{aligned}

Taking the limit of the outer exponent using the standard trigonometric limit $lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2}$ :

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

Therefore, combining the base and exponent limits:

\begin{aligned} \lim_{x\to 0} (\cos x)^{1/x^2} &= e^{-1/2} \\ &= \frac{1}{\sqrt{e}}. \end{aligned}

Quick Reference

The Number e

Computation of Certain Limits

Exercises

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