Limit of $x\left(\left(1 + \frac{1}{x}\right)^x - e\right)$ when $x\to\infty$

Question

I am stuck on how to calculate the following limit: $$\lim_{x\to\infty}x\left(\left(1 + \frac{1}{x}\right)^x - e\right).$$

Definitely, it has to be through l'Hôpital's rule. We know that $\lim_{x\to\infty} (1+(1/x))^x = e$. So, I wrote the above expression as $$\lim_{x\to\infty}\frac{\left(1 + \frac{1}{x}\right)^x - e}{1/x}.$$Both numerator and denominator tend to zero as x tends to infinity. I applied l'Hôpital's rule twice, but I got the limit equal to infinity which is clearly wrong. In the book, it says that the limit should be $-e/2$.

Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your l'Hôpital's rule to in both cases? Thanks a lot

Also, guys can you tell me how to write equations in this forum? Thanks.

Answer 1

One method is via the substitution $x = 1/y$. As $x \to \infty$, $y \to 0+$.

The given function can be written as $$ \begin{eqnarray*} \frac 1 y \left[ (1+y)^{1/y} - {\mathrm e} \right] &=& \frac{1}{y} \left[ \exp\left(\frac{\ln (1+y)}{y} \right) - \mathrm e \right] \\&=& \frac{\mathrm e}{y} \left[ \exp\left(\frac{\ln (1+y) - y}{y} \right) - 1 \right] \\ &=& \mathrm e \cdot \frac{\ln(1+y)-y}{y^2} \cdot \frac{ \exp\left(\frac{\ln (1+y) - y}{y} \right) - 1 }{\frac{\ln (1+y) - y}{y}} \end{eqnarray*} $$ Can you go from here? The last factor approaches $1$ by the standard limit $\frac{{\mathrm e}^z -1}{z} \to 1$ as $z \to 0$. The limit of the middle factor can be evaluated by L'Hôpital's rule.

Answer 2

You can indeed use l'Hôpital. Consider $$ f(x)=\left(1+\frac{1}{x}\right)^{\!x}=\exp\left(x\log\left(1+\frac{1}{x}\right)\right)=\exp(x\log(x+1)-x\log x) $$ Then $$ f'(x)=f(x)\left(\log(x+1)+\frac{x}{x+1}-\log x-1\right) $$ Then we have to compute $$ \lim_{x\to\infty}\frac{f'(x)}{-1/x^2} $$ Since the limit of $f(x)$ is $e$, we can concentrate on $$ \lim_{x\to\infty}\frac{\log(x+1)-\log x-\dfrac{1}{x+1}}{-1/x^2} $$ and use l'Hôpital again: $$ \lim_{x\to\infty}\frac{\dfrac{1}{x+1}-\dfrac{1}{x}+\dfrac{1}{(x+1)^2}}{2/x^3}= \lim_{x\to\infty}\frac{-x^3}{2x(x+1)^2}=-\frac{1}{2} $$ Reinstating the factor $e$, the limit is indeed $-e/2$.

Answer 3

I'll use the rule twice where you see a $\ast$, but this probably isn't the most elegant treatment:$$\begin{align}\lim_{x\to\infty}\frac{(1+1/x)^x-e}{1/x}&=\lim_{y\to0}\frac{(1+y)^{1/y}-e}{y}\\&\stackrel{\ast}{=}\lim_{y\to0}\frac{(1+y)^{1/y}\frac{d}{dy}\frac{\ln(1+y)}{y}}{1}\\&=\lim_{y\to0}(1+y)^{1/y}\frac{1-(1+y)\frac1y\ln(1+y)}{y(1+y)}\\&=e\lim_{y\to0}\frac{1-(1+y)\frac1y\ln(1+y)}{y(1+y)}\\&\stackrel{\ast}{=}e\lim_{y\to0}\frac{(\ln(1+y)-y)/y^2}{1+2y}\\&=e\frac{-1/2}{1+0}=-\frac{e}{2}.\end{align}$$Another solution that doesn't use the rule notes that$$(1+1/x)^x=\exp x\ln(1+\frac1x)=e\cdot\exp\left(-\frac{1}{2x}+o\left(\frac1x\right)\right)=e\left(1-\frac{1}{2x}+o\left(\frac1x\right)\right).$$Thus$$x((1+1/x)^x-e)=-\frac{e}{2}+o(1).$$

Answer 4

It's simpler to use Taylor formula at order $2$. Indeed, set $u=\frac1x$. The expression becomes $$x\biggl(\Bigl(1+\frac{1}{x}\Bigr)^x-\mathrm e\biggr)=\frac{(1+u)^{\tfrac1u}-\mathrm e}u=\frac{\mathrm e^{\tfrac{\ln(1+u)}u}-\mathrm e}u.$$

Now the numerator is \begin{align} \mathrm e^{\tfrac{\ln(1+u)}u}-\mathrm e&= \mathrm e^{\tfrac{u-\frac12 u^2 +o(u^2)}u}-\mathrm e=\mathrm e^{1-\frac{1}2u+o(u)}-\mathrm e= \mathrm e\bigl(\mathrm e^{\tfrac{u}2+o(u)}-1\bigr)\\ &=\mathrm e\Bigl(1+\frac u2+o(u)-1\Bigr)=\frac{\mathrm e\mkern2mu u}2+o(u)\sim_0 \frac{\mathrm e\mkern2mu u}2 \end{align} so that $$\frac{\mathrm e^{\tfrac{\ln(1+u)}u}-\mathrm e}u\sim_0 \frac{\mathrm e\mkern2mu u}{2u}=\frac{\mathrm e}2.$$