The Number *e* and Certain Limits

Limit	Value
$\underset{x\to 0}{lim}(1+x{)}^{1/x}$	$e$
$\underset{x\to +\mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{x}})}^x$	$e$
$\underset{x\to 0}{lim}{\displaystyle \frac{\ln (1+x)}{x}}$	$1$
$\underset{x\to 0}{lim}{\displaystyle \frac{e^x-1}{x}}$	$1$
$\underset{x\to 0}{lim}{\displaystyle \frac{(1+x{)}^r-1}{x}}$	$r$

The Number e

One of the most important limits in calculus is

$$\underset{x\to 0}{lim}(1+x{)}^{1/x}=e$$

where $e$ is an irrational number (like $\pi$) approximately equal to $2.718281828$.

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

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 $x\to \mathrm{\infty }$, $y$ approaches $1$, 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 ($u\to +\mathrm{\infty }$). Therefore,

$$\underset{x\to 0^{+}}{lim}(1+x{)}^{1/x}=\underset{u\to +\mathrm{\infty }}{lim}{(1+\frac{1}{u})}^u=e.$$

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

$$\underset{x\to +\mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=e$$

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Similarly, as $x\to 0^{-}$, $u\to -\mathrm{\infty }$. Therefore, $$\underset{x\to 0^{-}}{lim}(1+x{)}^{1/x}=\underset{u\to -\mathrm{\infty }}{lim}{(1+\frac{1}{u})}^u=e,$$ or equivalently $$\underset{x\to -\mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=e.$$

As $x\to \pm \mathrm{\infty }$, $y\to e\approx 2.71828$.

Computation of Certain Limits

Now that we know ${\displaystyle \underset{x\to 0}{lim}(1+x{)}^{1/x}=e}$, by applying the continuity of basic functions we can determine three significant limits needed for the next chapter:

$$\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}=1$$$$\underset{x\to 0}{lim}\frac{e^x-1}{x}=1$$$$\underset{x\to 0}{lim}\frac{(1+x{)}^r-1}{x}=r$$

Proof of the first limit. We observe that

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

It follows from the continuity of the logarithmic function that

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

Proof of the second limit. Let $e^x-1=u$, so $e^x=u+1$. Taking the natural log of both sides:

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

Therefore, we can rewrite the 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 \rightarrow 0} \frac{e^{x}-1}{x} &= \lim_{u \rightarrow 0} \frac{u}{\ln(1+u)}=\frac{1}{\displaystyle\lim_{u \rightarrow 0} \frac{\ln(1+u)}{u}} \\ &= \frac{1}{1}=1. \end{aligned}

Proof of the third limit. Let $(1+x{)}^r-1=u$, so $(1+x{)}^r=1+u$. Taking logarithms of both sides:

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

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)}\cdot\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,

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

Frequently Asked Questions

Why is the number $e$ important in calculus?

The number $e$ is the unique base for which the exponential function $e^x$ is its own derivative. This property makes $e$ the natural base for exponential and logarithmic functions throughout calculus and its applications in science and engineering. Every compound-growth and decay model involves $e$.

How is $\underset{x\to +\mathrm{\infty }}{lim}(1+1/x{)}^x=e$ related to compound interest?

If you invest $1$ at an annual interest rate of $100\%$ compounded $x$ times per year, after one year you have $(1+1/x{)}^x$ dollars. As the compounding frequency $x$ grows without bound (continuous compounding), the balance approaches $e\approx \mathrm{\$}2.718$. This is where $e$ first appeared historically.

Why does $\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}=1$ matter for derivatives?

This limit is precisely what you need to differentiate $\ln x$ from first principles. The derivative of $\ln x$ is $\frac{d}{dx}\ln x=\frac{1}{x}$, and the calculation reduces directly to evaluating $\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}=1$. Without this limit, the derivative of the natural logarithm cannot be established rigorously.

What is the general pattern connecting all three limits?

All three limits are consequences of $\underset{x\to 0}{lim}(1+x{)}^{1/x}=e$ combined with the continuity of logarithmic and exponential functions. The second limit follows from the first by substitution; the third generalizes the second to arbitrary real exponents $r$. Together they provide the toolkit for differentiating logarithmic, exponential, and power functions.