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Please give me some hints or solution for this limit. I had it on my exam and I'm curious how to solve it. $$ \lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x} $$

dustin
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3 Answers3

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Easiest way is by series expansion: $$\frac{1}{x}\log(1+x) = \frac{1}{x} \left( x - \frac{x^2}{2} + O(x^3) \right) = \left( 1 - \frac{x}{2} + O(x^2) \right) $$ $$ (x+1)^{1/x} = e^{\frac{1}{x}\log(1+x) } = e^{\left( 1 - \frac{x}{2} + O(x^2) \right)} = \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n}{n!} $$ $$ (x+1)^{1/x}-e = \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n}{n!} - \sum_{n=0}^\infty \frac{1}{n!} = \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n-1}{n!} $$ $$ \frac{(x+1)^{1/x}-e}{x} = \sum_{n=0}^\infty \frac{\left( 1 - n\frac{x}{2} + O(x^2) \right)-1}{xn!} = \sum_{n=0}^\infty \frac{ - n\frac{x}{2} + O(x^2) }{xn!} $$ $$ \frac{(x+1)^{1/x}-e}{x} = -\frac{1}{2} \sum_{n=0}^\infty \frac{n + O(x)}{n!} = -\frac{1}{2} \sum_{n=0}^\infty \frac{n }{n!} +O(x) $$ $$ \frac{(x+1)^{1/x}-e}{x} = -\frac{1}{2} \sum_{n=1}^\infty \frac{n}{n!} +O(x) = -\frac{1}{2} \sum_{m=0}^\infty \frac{1}{(m+1)!} +O(x) = -\frac{e}{2} + O(x) $$ from which it follows that $$ \lim_{x\rightarrow \infty}\frac{(x+1)^{1/x}-e}{x} = -\frac{e}{2} $$

Mark Fischler
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  • Could you please describe how you get from $\left( 1-\frac{x}{2}+O(x^2)\right)^n$ to $1-n\frac{x}{2} + O(x^2)$? – imourad Jan 13 '15 at 01:13
  • Do you really think this is "the easiest way"? Easier than, say l'Hospital? It is a nice way, though...but easiest? – Timbuc Jan 13 '15 at 03:32
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Hint: Consider the function $f(x) = (x + 1)^{1/x}$ for $x \neq 0$ and $f(0) = e$. Then $f$ is right-continuous at everywhere and the limit you are asked to find is the right-derivative of $f$ at $x = 0$, if it exists. So find $f'$ and evaluate it at $x = 0$.

Simon S
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  • http://www.wolframalpha.com/input/?i=%28%281%2Bx%29^%281%2Fx%29%29%27

    It can't be evaluated at 0, or am I wrong?

     – Hasan Saad Jan 12 '15 at 21:45
  • He explicitly defined the function such $$f(x) = \begin{cases} (x+1)^{1/x}&\text{ if }x\neq 0\ e&\text{ if } x=0 \end{cases} $$ – Eff Jan 12 '15 at 21:46
  • Oops, sorry. I didn't notice. – Hasan Saad Jan 12 '15 at 21:49
  • However it turns out the form of the derivative is ugly and requires repeated applications of l'Hopital. Probably better to do what Timbuc has done instead and use l'Hopital on the original expression. – Simon S Jan 12 '15 at 21:50
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Use l'Hospital, since $\;(x+1)^{1/x}\xrightarrow[x\to 0]{}e\;$ , writing

$$(x+1)^{1/x}=e^{\frac1x\log(x+1)}\implies\;\text{the limit thus is} \;\;\lim_{x\to 0}\frac{e^{\frac1x\log(x+1)}-e}x\;:$$

$$\lim_{x\to 0}\frac{e^{\frac{\log(x+1)}x}-e}x\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{x-(x+1)\log(x+1)}{x^2(x+1)}e^{\frac{\log(x+1)}x}\stackrel{\text{l'H}}=-\frac e2$$

Timbuc
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  • Thanks. At my exam I managed to get to the last equation before the result, but I have no idea how you get $-\frac{e}{2}$. –  Jan 12 '15 at 21:53
  • @EricTroy $$e^{\frac{\log(x+1)}x}\xrightarrow[x\to 0]{}e$$and $$\frac{x-(x+1)\log(x+1)}{x^2(x+1)}\xrightarrow[x\to 0]{}-\frac12$$ both by l'Hospital and continuity of the exponential function – Timbuc Jan 12 '15 at 21:56