Evaluate $\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x$ I tried:
$$\left(\frac{x}{1+x}\right)^x=A$$
$$\ln(A)=x\ln\left(\frac{x}{1+x}\right)$$
$$\lim_{x\to 1^{+}}\frac{\ln x-\ln (x+1)}{\frac1x}=\lim_{x\to 1^{+}}\frac{\frac1x-\frac1{x+1}}{\frac{-1}{x^2}}=\frac{-1}{2}$$
I applied L'Hopital rule in last step.
I get $\lim(A)_{x\to 1^{+}}=e^{\frac{-1}{2}}$
But the final answer provided in the book is $e^{-1}$. where I made a mistake?
 A: Hint
You cannot apply l'Hopital here since it's not indeterminated. Nevertheless, $$\lim_{x\to 1^+}\frac{\ln(x)-\ln(1+x)}{1/x}=-\ln(2).$$
I let you conclude (and the limit won't be $e^{-1}$ but $\frac{1}{2}$).
A: If our function is wrote in this form, you can't use  l'Hopital.
Since, $e^x$ and $\ln$ are continuous functions we obtain:
$$\lim_{x\rightarrow1}\left(\frac{x}{1+x}\right)^x=\lim_{x\rightarrow1}e^{x\ln\frac{x}{1+x}}=e^{\lim\limits_{x\rightarrow1}x\ln\frac{x}{1+x}}=e^{1\cdot\ln\lim\limits_{x\rightarrow1}\frac{x}{1+x}}=e^{\ln\frac{1}{2}}=\frac{1}{2}.$$
A: As already noticed by continuity
$$\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x=\frac12$$
it seems that, fixed the typo, the original question is referring to
$$\lim_{x\to 0^{+}}\left(\frac{x}{1+x}\right)^x=\lim_{x\to 0^{+}}\frac{x^x}{(1+x)^x}=1$$
or to
$$\lim_{x\to \infty}\left(\frac{x}{1+x}\right)^x=\lim_{x\to \infty}\left(1-\frac{1}{1+x}\right)^x=\frac1e$$
note that in any case we don't need l'Hospital's rule to obtain the result.
