Applying L'Hopital's rule for $\lim_{x\to\infty} x((1+1/x)^x-e)$

Question

I know the limit is $-e/2$ but I can't get there. I know I should be using L'Hopitals Rule here, I tried both $0/0$ and $\infty/\infty$, either way it's a big mess. Please help. Maybe you can use the $e^{log(...)}$ trick but I haven't found it to be useful.

Edit: I'm not familiar with the Big-O notation

Answer 1

HINT: If one wishes to use L'Hospital's Rule

Letting $t=1/x$ we can write

$$\begin{align} \lim_{x\to\infty}x\left(\left(1+\frac1x\right)^x-e\right)&=\lim_{t\to 0}\frac1t\left(\left(1+t\right)^{1/t}-e\right)\\\\ &\overbrace{=}^{LHR}\lim_{t\to0}\frac{d(1+t)^{1/t}}{dt} \end{align}$$

Apply LHR two more times to evaluate the limit of the derivative.

Can you finish?

Answer 2

As an alternative to the nice solution with l'Hopital given by Mark Viola, by Taylor's expansion for $t\to 0$

• $\log (1+t)=t-\frac12t^2 +o(t^2)$

• $e^t=t+o(t)$

we have

$$\left(1+\frac1x\right)^x=e^{x\log\left(1+\frac1x\right)}=e^{x\left(\frac1x-\frac1{2x^2}+o(1/x^2)\right)}=e^{1-\frac1{2x}+o(1/x)}=e\cdot e^{-\frac1{2x}+o(1/x)}=e\left(1-\frac1{2x}+o(1/x)\right)$$

and therefore

$$x\left(\left(1+\frac1x\right)^x-e\right)=x\left(e-\frac e{2x}+o(1/x)-e\right)=-\frac e{2}+o(1)\to -\frac e 2$$

Answer 3

It is easier to do if you let $x=n$ an integer. $f(n)=(!+\frac{1}{n})^n=\sum_{k=0}^n \binom{n}{k}(\frac{1}{n})^k$. $e=\sum_{k=0}^\infty\frac{1}{k!}$. Expand $\binom{n}{k}$ as powers of $n$ (note) and keep the first two terms to get $\frac{1}{k!}(n^k-\frac{k(k-1)n^{k-1}}{2})$. Use this as an approximation to $f(n)$ and get $n(f(n)-e)\approx -\sum_{k=2}^n\frac{1}{2(k-2)!}\to -\frac{e}{2}$, ignoring higher order terms in the expansion of $e$.

Note: Dropping the remaining terms is justified since dividing by $n^k$ and multiplying by $n$ leaves powers of $\frac{1}{n}$ which $\to 0$ as $n\to \infty$.