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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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    "Definitely, it has to be through l'Hôpital's rule" **Well!**2015-09-30

4 Answers 4

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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.

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    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 lot2011-10-17