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In Problems in Mathematical Analysis, Volume 2 by Kaczor and Nowak, Problem 2.3.34 is stated as follow.

Find this limit using L`Hôpital rule


I have tried by using L'Hospital rule five times but have no result. I can compute this limit by using Taylor expansion.


marked as duplicate by YuiTo Cheng, Lee David Chung Lin, max_zorn, José Carlos Santos calculus Jul 4 at 8:02

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  • $\begingroup$ Did you write it as $((1+1/x)^x - e)/(1/x)$? Applying l'Hop once to that should get you where you need to go. $\endgroup$ – Simon S Mar 19 '15 at 15:24
  • $\begingroup$ yes, I did. But it seems more complicated. $\endgroup$ – Jean Piere Peter Mar 19 '15 at 15:26
  • $\begingroup$ Write out your derivative for the numerator and denominator $\endgroup$ – Simon S Mar 19 '15 at 15:28
  • $\begingroup$ I agree with @Simon S. Except in trivial cases, it is in fact almost never simpler to use L'Hopital's rule than to use perturbation expansions, if you now what you are doing with the expansions. $\endgroup$ – Mark Fischler Mar 19 '15 at 15:29
  • $\begingroup$ It's upperbounded by 0 and monotone increasing. The limit should be 0. $\endgroup$ – Alex Mar 19 '15 at 15:31

First, change the limit into this form,

$$\lim\limits_{x\to 0}\frac{(1+x)^{1/x}-e}{x}.$$

and apply L`Hôpital rule to get,

$$\lim\limits_{x\to 0} \;(x+1)^{1/x} \frac{\left(\frac{x}{x+1}-\log (x+1)\right)}{x^2}.$$

Since $\lim\limits_{x\to 0} \;(x+1)^{1/x} =e$ , you may want to know if $\lim\limits_{x\to 0}\frac{\left(\frac{x}{x+1}-\log (x+1)\right)}{x^2}$ exist. Apply L`Hôpital rule again, $$\lim\limits_{x\to 0}\frac{\left(\frac{x}{x+1}-\log (x+1)\right)}{x^2} = \lim\limits_{x\to 0} \frac{-1}{2 (x+1)^2} = -\frac{1}{2}$$

It exist and we are done,

$$\lim\limits_{x\to 0}\frac{(1+x)^{1/x}-e}{x}=\lim\limits_{x\to 0} \;(x+1)^{1/x} \lim\limits_{x\to 0}\frac{\left(\frac{x}{x+1}-\log (x+1)\right)}{x^2} = -\frac{e}{2}.$$


You can use L'Hopital Rule and substitution to get the answer. In fact \begin{eqnarray} \lim_{x\to\infty}x\left[(1+\frac 1x)^x-e\right]&=&\lim_{x\to\infty}\frac{(1+\frac 1x)^x-e}{\frac 1x} \quad (\text{let } u=\frac 1x)\\ &=&\lim_{u\to 0}\frac{(1+u)^{\frac{1}{u}}-e}{u}\\ &=&\lim_{u\to 0}\frac{e^{\frac{1}{u}\ln(1+u)}-e}{u}\\ &=&\lim_{u\to 0}e(\frac{1}{u}\ln(1+u))'\\ &=&-\frac{e}{2}. \end{eqnarray} Here we use $\ln(1+x)=x-\frac{x^2}{2}+o(x^2)$ to compute $$ \lim_{u\to 0}(\frac{1}{u}\ln(1+u))'=-\frac 12. $$


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