# What is the limit as x approaches infinity of e^x?

Aug 23, 2017

Another perspective...

#### Explanation:

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As a Real function

Treating ${e}^{x}$ as a function of Real values of $x$, it has the following properties:

• The domain of ${e}^{x}$ is the whole of $\mathbb{R}$.

• The range of ${e}^{x}$ is $\left(0 , \infty\right)$.

• ${e}^{x}$ is continuous on the whole of $\mathbb{R}$ and infinitely differentiable, with $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$.

• ${e}^{x}$ is one to one, so has a well defined inverse function ($\ln x$) from $\left(0 , \infty\right)$ onto $\mathbb{R}$.

• ${\lim}_{x \to + \infty} {e}^{x} = + \infty$

• ${\lim}_{x \to - \infty} {e}^{x} = 0$

At first sight this answers the question, but what about Complex values of $x$?

$\textcolor{w h i t e}{}$
As a Complex function

Treated as a function of Complex values of $x$, ${e}^{x}$ has the properties:

• The domain of ${e}^{x}$ is the whole of $\mathbb{C}$.

• The range of ${e}^{x}$ is $\mathbb{C} \text{\} \left\{0\right\}$.

• ${e}^{x}$ is continuous on the whole of $\mathbb{C}$ and infinitely differentiable, with $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$.

• ${e}^{x}$ is many to one, so has no inverse function. The definition of $\ln x$ can be extended to a function from $\mathbb{C} \text{\} \left\{0\right\}$ into $\mathbb{C}$, typically onto $\left\{x + i y : x \in \mathbb{R} , y \in \left(- \pi , \pi\right]\right\}$.

What do we mean by the limit of ${e}^{x}$ as $x \to \text{infinity}$ in this context?

From the origin, we can head off towards "infinity" in all sorts of ways.

For example, if we just set off along the imaginary axis, the value of ${e}^{x}$ just goes round and around the unit circle.

If we choose any complex number $c = r \left(\cos \theta + i \sin \theta\right)$, then following the line $\ln r + i t$ for $t \in \mathbb{R}$ as $t \to + \infty$, the value of ${e}^{\ln r + i t}$ will take the value $c$ infinitely many times.

We can project the Complex plane onto a sphere called the Riemann sphere ${\mathbb{C}}_{\infty}$, with an additional point called $\infty$. This allows us to picture the "neighbourhood of $\infty$" and think about the behaviour of the function ${e}^{x}$ there.

From our preceding observations, ${e}^{x}$ takes every non-zero complex value infinitely many times in any arbitrarily small neighbourhood of $\infty$. That is called an essential singularity at infinity.

Aug 23, 2017

Exaplanation using logarithms.

#### Explanation:

The limit does not exist because as $x$ increases without bond, ${e}^{x}$ also increases without bound. ${\lim}_{x \rightarrow \infty} {e}^{x} = \infty$.

Te xplanation of why will depand a great deal on the definitions of ${e}^{x}$ and $\ln x$ with which you are working.

I like to define $\ln x = {\int}_{1}^{x} \frac{1}{t} \mathrm{dt}$ for $x > 0$, then prove that $\ln x$ is invertible (has an inverse) and define ${e}^{x}$ as the inverse of $\ln x$.

Since $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$, it is clear that $\ln x$ is increasing

In the process, I also prove to my students that that ${\lim}_{x \rightarrow 00} \ln x = \infty$.
(The proof uses ideas similar to those used in showing that the harmonic series diverges.)

Knowing these things allows us to reason as follows.

For any positive Real number $T$, $\ln T$ is defined and for all $x \ge \ln T$, we have ${e}^{x} \ge {e}^{\ln T} = T$.

That is, for every $T$, there is an $M$ (namely $\ln T$ such that if $x \ge M , t h e n$f(x) >= T

Therefore, by definition, ${\lim}_{x \rightarrow \infty} {e}^{x} = \infty$