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

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@UnadulteratedImagination "Should be"? How can the answer depend on $x$ when we are asked to take the limit $x \to 0$?! –  Fly by Night Jan 29 '13 at 20:21
    
$e^2$ using log and L'Hospital –  UnadulteratedImagination Jan 29 '13 at 20:37
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6 Answers

up vote 12 down vote accepted

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

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It took me 1 month to finally understand L'Hospital rule. Thanks! –  bryansis2010 Feb 26 '13 at 15:45
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Using L'Hôpital's Rule, $$ \begin{eqnarray} \lim_{x \rightarrow 0} \left(e^x+x\right)^{1/x} &=& \exp\left\{\lim_{x \rightarrow 0} \ \ln \left[\left(e^x+x\right)^{1/x}\right]\right\} \\ &=& \exp\left[\lim_{x \rightarrow 0} \frac{\ln \left(e^x+x\right)}{x}\right] \\ &=& \exp\left(\lim_{x \rightarrow 0} \frac{e^x+1}{e^x+x}\right) \\ &=& e^2 \end{eqnarray} $$

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The jump from the second line to the third one requires, I think, a thorough justification –  DonAntonio Jan 29 '13 at 20:28
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$$ \large \begin{align*} \lim_{x \to 0}\ (e^x + x)^ \frac {1} {x} &= e \; \lim_{x \to 0}\ \left (1 + \frac{x}{e^x} \right )^ {\frac {1} {x}}\\ &= e \; \lim_{x \to 0}\ \left (\left (1 + \frac{x}{e^x} \right )^ {\frac {1}{\frac x{e^x}}} \right)^{\frac{1}{e^x}}\\ &= e e = e^2\\ \end{align*} $$

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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 $e^2$.

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Nice observation! +1 –  amWhy Jan 30 '13 at 0:44
    
@amWhy: Above fact has helped me a lot. –  B. S. Jan 30 '13 at 6:45
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\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 $\lim_{x \to 0} (e^x+x)^{\frac{1}{x}} = e^2$.

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$$\begin{align}\lim\limits_{x \to 0}\ (e^x + x)^ \frac {1} {x} &=\lim\limits_{x \to 0}\ (1+x+\frac{x^2}{2!}+\dots + x)^ \frac {1} {x}\\ &=\lim\limits_{x \to 0}\ (1+2x+\frac{x^2}{2!}+\dots )^ \frac {1} {x}=e^2\end{align}$$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.

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You need to explain why the final limit is $e^2$. –  Fly by Night Jan 29 '13 at 20:23
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This doesn't convince me. –  Patrick Li Jan 29 '13 at 20:29
    
@PatrickLi : I think you will be convinced with the second method that I just add with my answer. –  A.D Jan 30 '13 at 3:38
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