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.
Best Answer
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.