this is my first post. I've been struggling whit this limit for too long (without using l'Hôpital's rule):
$$\lim_{x\to {\infty}} \left(\frac{x-1}{x+1}\right)^x$$
My answer is $\frac1e$, but the correct answer should be $\frac{1}{e^2}$. Could anyone help me understand why? Thanks in advance!
$\endgroup$3 Answers
$\begingroup$This is $\lim_{x\to\infty}\left(A(x)/B(x)\right)$ where$$A(x)=\left(1-\frac1x\right)^x$$and$$B(x)=\left(1+\frac1x\right)^x.$$You have probably seen that $B(x)\to e$ and $A(x)\to e^{-1}$ as $x\to\infty$.
$\endgroup$ 6 $\begingroup$HINT
Use that
$$\left(\frac{x-1}{x+1}\right)^x=\left[\left(1+\frac{-2}{x+1}\right)^{x+1}\right]^{\frac{x}{x+1}}$$
Refer also to the earlier related
(and you can find many other similar examples on MSE)
$\endgroup$ $\begingroup$You can go to exponential : for all $x\in \mathbb R_+^*$\begin{eqnarray} \left(\frac{x-1}{x+1}\right)^x&=&\exp\left(x \ln\left(\frac{x-1}{x+1}\right)\right)\\ &=& \exp \left( x \ln\left(1-\frac{2}{x+1}\right)\right)\\ &=& \exp\left(x\left(-\frac{2}{x+1}+O(1/x^2)\right)\right)\\ &=& \exp(-2+o(1))\\ &=& \frac{1}{e^2}+o(1) \end{eqnarray}
$\endgroup$ 4
