Find the limit of $\dfrac{\frac{1}{e}(1+x)^{1/x}-1+\frac{x}{2}}{x^2}$ when $x\to0$.
I tried applying L'Hospital rule, but it is not working here. How should I solve this?
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2Taylor expansion of the exponential. – Daniel Fischer Oct 06 '15 at 08:52
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I expanded using Taylor also,but i could not reach the answer. – user1442 Oct 06 '15 at 08:54
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1What did you get with the Taylor expansion? – Daniel Fischer Oct 06 '15 at 08:57
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@DanielFischer, Why did you suggest Taylor expansion of the exponential? I can't see an $e^x$ in this expression. Just a $\frac{1}{e}$. What have I missed? – R_D Oct 06 '15 at 08:59
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2@Rise $e^{-1}(1+x)^{1/x}=\exp\left(\frac1x\log(1+x)-1\right)$. – Did Oct 06 '15 at 09:00
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Of course! Thank you @Did – R_D Oct 06 '15 at 09:01
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I got the answer using the suggestion given by Did and Daniel Fischer,thank you. – user1442 Oct 06 '15 at 09:10
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2@user1442 Great. Will you write an answer? – Daniel Fischer Oct 06 '15 at 09:10
1 Answers
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With the definition of $a^b$ we have for the numerator
\begin{align*} (1+x)^\frac{1}{x} \cdot \exp(-1) -1+\frac{x}{2}&= \exp\left(\frac{1}{x} \cdot \ln(1+x) \right) \cdot \exp(-1) -1+\frac{x}{2}\\ &=1-\frac{x}{2}+\frac{11 x^2}{24}-1+\frac{x}{2} +\mathcal{O}(x^3)\\ &= \frac{11x^2}{24} +\mathcal{O}(x^3) \end{align*}
Now the limit is equal to $$\lim_{x\to 0} \frac{\frac{11}{24} x^2}{x^2}=\frac{11}{24}$$
Maybe some words how to derive they taylor series without calculating to much. We have $$\log(1+x)= \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} = x-\frac{x^2}{2} \pm \dots$$ and $$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2}+\dots$$.
So we have \begin{align*} \exp( \log(1+x)\cdot x^{-1} -1)&= 1+ \log(1+x)\cdot x^{-1} -1 + \frac{(\log(1+x)\cdot x^{-1} -1)^2}{2} +\mathcal{O}((\log(1+x)\cdot x^{-1} -1)^3)\\ &= 1-\frac{x}{2} +\frac{x^2}{3} + \left(\frac{-x}{2}+\frac{x^2}{3}\right)^2 \cdot \frac{1}{2}+ \mathcal{O}(x^3) \end{align*}
Dominic Michaelis
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