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

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.

• Equations are written as standard LaTeX. Check out this meta question Oct 17 '11 at 4:16
• Perhaps you can show your applications of LHopital's? (I realize that since you didn't know how to write equations, that would have been difficult; one possibility is to scan your pages: someone will come along and type the equations). Oct 17 '11 at 4:23
• You could also use Maclaurin expansion, if you're familiar with that. Oct 17 '11 at 4:30
• After applying the L'Hopspital's rule for the first time, we get: Oct 17 '11 at 4:42
• When you apply l'Hospital rule for the first time you get \lim e \frac{ln(1+1/x)-1/1+x)}{\frac{1-}{x^2}}. When you apply it for the second time you get: lim e \frac{\frac{1}{1+x}-\frac{1}{x}+\frac{1}{(1+x)^2}}{\frac{2}{x^3}} which definitely goes to infinity. Where is the mistake? Oct 17 '11 at 4:54

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.

• When you apply l'Hospital rule for the first time you get \lim e \frac{ln(1+1/x)-1/1+x)}{\frac{1-}{x^2}}. When you apply it for the second time you get: lim e \frac{\frac{1}{1+x}-\frac{1}{x}+\frac{1}{(1+x)^2}}{\frac{2}{x^3}} which definitely goes to infinity. Where is the mistake? Oct 17 '11 at 4:51
• @Zi2018 I did not try applying L'Hospital's rule to the given function directly since that seemed too messy. If I get something sensible, I will give the answer in a comment later on. Oct 17 '11 at 4:54
• Thanks a lot Srivatsan Narayanan. I tried your method and it worked. Indeed the limit is -e/2. But, I can't figure out what wrong with my method is, although they look the same to me except for the change of variable. Can anyone try my method and tell me where my mistake is? Thanks a lot Oct 17 '11 at 5:04

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 $$-e/2$$.

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$$x((1+1/x)^x-e)=-\frac{e}{2}+o(1).$$

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 $$\frac{\mathrm e^{\tfrac{\ln(1+u)}u}-\mathrm e}u\sim_0 \frac{\mathrm e\mkern2mu u}{2u}=\frac{\mathrm e}2.$$