Limit of a Function: $ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}$

Question

What is the limit of the following function which consists of an exponential and algebraic expression? $$ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}\;\;?$$

Answer 1

First off, take the logarithm and look at $\displaystyle\lim_{x \to 0} \frac{\ln(e^x + x)}{x}$.

Apply L'Hopital's rule to get $\displaystyle\lim_{x \to 0} \frac{\ln(e^x + x)}{x} = \lim_{x \to 0} \frac{e^x + 1}{e^x + x} = 2$.

Hence $\displaystyle\lim_{x \to 0}(e^x + x)^\frac{1}{x} = \exp(\lim_{x \to 0} \frac{\ln(e^x + x)}{x}) = \exp(2) = e^2$.

Answer 2

Using L'Hôpital's Rule, $$ \begin{eqnarray} \lim_{x \rightarrow 0} \left(e^x+x\right)^{1/x} &=& \exp\left\{\lim_{x \rightarrow 0} \ \ln \left[\left(e^x+x\right)^{1/x}\right]\right\} \\ &=& \exp\left[\lim_{x \rightarrow 0} \frac{\ln \left(e^x+x\right)}{x}\right] \\ &=& \exp\left(\lim_{x \rightarrow 0} \frac{e^x+1}{e^x+x}\right) \\ &=& e^2 \end{eqnarray} $$

Answer 3

$$ \large \begin{align*} \lim_{x \to 0}\ (e^x + x)^ \frac {1} {x} &= e \; \lim_{x \to 0}\ \left (1 + \frac{x}{e^x} \right )^ {\frac {1} {x}}\\ &= e \; \lim_{x \to 0}\ \left (\left (1 + \frac{x}{e^x} \right )^ {\frac {1}{\frac x{e^x}}} \right)^{\frac{1}{e^x}}\\ &= e e = e^2\\ \end{align*} $$

Answer 4

Note that when $\alpha(x)$ is so small when $x\to\infty$, then we have $$a^{\alpha(x)}-1\sim\alpha(x)\ln(a)$$ Use this fact and set $a=e, \alpha(x)=x$, so have the limit $e^2$.

Answer 5

\begin{eqnarray} \lim_{x \to 0} (e^x+x)^{\frac{1}{x}} &=& \lim_{x \to 0} e^{\frac{1}{x} \ln (e^x + x)}\\ &=& \lim_{x \to 0} e^{\frac{1}{x} \ln (e^x(1 + xe^{-1}))} \\ &=& \lim_{x \to 0} e^{\frac{1}{x} \ln e^x} e^{\frac{1}{x} \ln (1 + xe^{-x})}\\ &=& e \lim_{x \to 0} e^{\frac{1}{x} \ln (1 + xe^{-x})} \end{eqnarray} Since $y \mapsto \ln(1+y)$ is differentiable at zero, for any $\epsilon>0$, there is a $\delta>0$ such that if $|y|<\delta$, then $|\ln(1+y)-y|\leq \epsilon |y|$. This gives $|\ln (1 + xe^{-x})-x e^{-x}| \leq \epsilon |x| e^{-x}$, or equivalently, $|\frac{1}{x}\ln (1 + xe^{-x})-e^{-x}| \leq \epsilon e^{-x}$, from which it follows that $\lim_{x \to 0} \frac{1}{x}\ln (1 + xe^{-x}) = 1$, from which we obtain $\lim_{x \to 0} (e^x+x)^{\frac{1}{x}} = e^2$.

Answer 6

$$\begin{align}\lim\limits_{x \to 0}\ (e^x + x)^ \frac {1} {x} &=\lim\limits_{x \to 0}\ (1+x+\frac{x^2}{2!}+\dots + x)^ \frac {1} {x}\\ &=\lim\limits_{x \to 0}\ (1+2x+\frac{x^2}{2!}+\dots )^ \frac {1} {x}=e^2\end{align}$$Note that since $x \to 0$ we can assume $\frac{x^2}{2!}+ \frac{x^3}{3!}+\cdots =0$ and $$\lim_{x \to 0}(1+2x)^{\frac{1}{x}}=\lim_{u \to 0}(1+u)^{\frac{1}{u}{2}}=e^2$$ where $u=2x.$

Alternatively note that $$\lim_{x \to 0}(1+2x+\frac{x^2}{2!}+\dots )^{\frac{1}{x}}=\lim_{u \to 0}(1+u)^{\frac{1}{u}[\lim_{x \to 0}\frac{u}{x}]} = \exp{\lim_{x \to 0}\frac{u}{x}}=e^2$$ where $u=2x+\frac{x^2}{2!}+\cdots$ and

$$\begin{align}\lim_{x \to 0}\frac{u}{x} &=\lim_{x \to 0}\frac{2x+\frac{x^2}{2!}+\cdots}{x}\\ &=\lim_{x \to 0}[2+\frac{x}{2!}+\frac{x^2}{3!}+\cdots]\\ &= 2\end{align}$$

I personally like the first method because with some logical fact the given limit reduces to a smooth limit. Some people dislike the first method though it is not wrong. If anybody want to avoid the logical fact and wanna solve the problem only with mathematical tools he/she may solve the problem in the second method.