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The question is to find $\lim\limits_{n\to\infty}{n\left(\left(1+\frac{1}{n}\right)^n-e\right)}$.

I am trying to use L'Hospital Rule, then I have $$\lim_{n\to\infty}{\frac{(1+\frac{1}{n})^n-e}{\frac{1}{n}}}=\lim_{n\to\infty}\frac{{n(1+\frac{1}{n})^{n-1}*(-\frac{1}{n^2})}}{{-\frac{1}{n^2}}}=\lim_{n\to\infty}{n\left(1+\frac{1}{n}\right)^{n-1}}=\infty.$$

I feel I made some mistakes somewhere, and I appreciate anyone can point out for me. Or if my method is wrong, I hope to get some hint on how to solve this one.

Michael Rozenberg
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Runj
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    You are missing a left parenthesis in the problem. Is it supposed to go after the first $x$? – Ross Millikan Jun 20 '17 at 19:51
  • Yes, I just edit it. Thanks. – Runj Jun 20 '17 at 19:54
  • The derivative of $(1 + x^{-1})^x$ is $$ - \frac{x}{x^2}\left(1 + \frac1x \right)^{x - 1} + \left(1 + \frac1x \right)^x \log \left(1 + \frac1x \right).$$ – Trevor Gunn Jun 20 '17 at 19:58
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    The derivative on the top is incorrect: IIRC $\frac{d}{dx}(f(x)^{g(x)}) = g(x) f(x)^{g(x)-1} f'(x) + f(x)^{g(x)} (\ln f(x)) g'(x)$. – Daniel Schepler Jun 20 '17 at 19:58
  • There is a solution exploiting the Stoltz-Cesaro theorem and the fact that (by the AM-GM inequality) $\left{\left(1+\frac{1}{n}\right)^n\right}{n\geq 1}$ is increasing while $\left{\left(1+\frac{1}{n}\right)^{n+1/2}\right}{n\geq 1}$ is decreasing, but I am too lazy to write it. – Jack D'Aurizio Jun 20 '17 at 20:35

3 Answers3

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The hint: Use $\ln$. It would help! $$\lim_{n\rightarrow+\infty}n\left(\left(1+\frac{1}{n}\right)^n-e\right)=\lim_{x\rightarrow{0^+}}\frac{(1+x)^{\frac{1}{x}}-e}{x}=$$ $$=\lim_{x\rightarrow{0^+}}\frac{\left((1+x)^{\frac{1}{x}}-e\right)'}{x'}=\lim_{x\rightarrow0^{+}}\left(e^{{\frac{\ln{(1+x)}}{x}}}\right)'=$$ $$=e\lim_{x\rightarrow0^{+}}\left(\frac{\ln{(1+x)}}{x}\right)'=e\lim_{x\rightarrow0^+}\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}=$$ $$=e\lim_{x\rightarrow0^+}\frac{x-(1+x)\ln(1+x)}{x^3+x^2}=e\lim_{x\rightarrow0^+}\frac{-\ln(1+x)}{3x^2+2x}=-\frac{e}{2}.$$

Michael Rozenberg
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The error in your answer:

The derivative of $\left(1+\frac 1x\right)^x$ is not $\left(1+\frac 1x\right)^{x-1}\frac{-1}{x^2}$.

It is true that the derivative of $\left(1+\frac1x\right)^{n}$ is $\left(1+\frac1x\right)^{n-1}\frac{-1}{x^2},$ it is not true if you replace $n$ with $x$.

If we write $h(x)=\left(1+\frac1x\right)^x=e^{x\log\left(1+\frac1x\right)}$ then $$\begin{align}h'(x)&=h(x)\left[\log\left(1+\frac1x\right)+\frac{x}{1+\frac{1}{x}}\frac{-1}{x^2}\right]\\ &=h(x)\left[\log\left(1+\frac1x\right) -\frac{1}{1+x}\right] \end{align}$$

So I'm not sure L'Hopital means you want to compute:

$$\lim_{x\to\infty} h(x)\frac{\log \left(1+\frac1x\right)-\frac{1}{1+x}}{\frac{-1}{x^2}}=e\lim_{x\to\infty} x^2\left(\log \left(1+\frac1x\right)-\frac{1}{1+x}\right)$$

Not sure if that is any easier.

Thomas Andrews
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You can actually do it without l'Hopital's rule, too. Define $f(x)=(1+x)^{1/x}$. As the other answers, and Michael Rozenberg have pointed out, $\lim_{n\to+\infty}n\left((1+\frac1n)^n-e\right)=\lim_{x\to0^+}\frac{(1+x)^{1/x}-e}{x}$. As $\lim_{x\to0^+}(1+x)^{1/x}=e$, as shown here, your limit equals $f'(x)$ as $x\to0$.

By setting $\ln f(x)=\frac{1}{x}\ln(1+x)$ and differentiating both sides with respect to $x$, we can find that $f'(x)=f(x)\left(\frac{x-(1+x)\ln (1+x)}{x^2(1+x)}\right)$. This gives the following:

$$f'(x)=\underbrace{\left(1+x\right)^{\frac{1-x}{x}}}_t\left(\underbrace{\frac{x-\ln \left(1+x\right)}{x^2}}_u-\underbrace{\frac{\ln \left(1+x\right)}{x}}_v\right)$$

As $x\to0$, the limit of $t$ is $\lim \left((1+x)^{1/x}\right)^{1-x}=\left(e\right)^{1-0}$. The limit of $u$ is $\lim\frac{x-\left(x-x^2/2+\mathcal{O}(x^3)\right)}{x^2}=\frac12+\frac{\mathcal{O}(x^3)}{x^2}$, by the Taylor series for $\ln(1+x)$, as shown here. By the same reasoning, the limit of $v$ is $\lim\frac{x-\mathcal{O}(x^2)}{x}=1$. Hence, $\lim_{x\to0}f'(x)=e(\frac{1}{2}-1)$.

Jam
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