# L'Hôpital's rule in real analysis

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.

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Equations are written as standard LaTeX. Check out this meta question –  Arturo Magidin 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). –  Arturo Magidin Oct 17 '11 at 4:23
You could also use Maclaurin expansion, if you're familiar with that. –  Hans Lundmark Oct 17 '11 at 4:30
After applying the L'Hopspital's rule for the first time, we get: –  M.Krov 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? –  M.Krov 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.