I am watching some videos of limit exercises, and here is a formula I don't get it. $$\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}\right)^n=\exp\left({\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-1\right)n}\right)$$ It has a $-1$ on RHS. I thought the LHS should equal to $\lim_{n\rightarrow\infty} \exp\left({\ln(\frac{n}{n+1})n}\right)=\exp\left(\lim_{n\rightarrow\infty}\ln\left(\frac{n}{n+1}\right)n\right)$. However, this is wrong after I check it on wolframalpha, since the final value should be $1/e$ and $\exp\left(\lim_{n\rightarrow\infty}\ln\left(\frac{n}{n+1}\right)n\right)$ give you infinity.
I think I can't shift the limit into exponent, but I thought $\lim_{x→c}f(g(x))=f(\lim_{x→c}g(x))$ hold if $f(x)$ is continuous, and $e^x$ is continuous.
I am wondering where this -1 comes from, and where I made mistake? Any comment will be appreciated!
As noticed in the comments you are right, indeed we are using that $A^B = e^{B\log A}$ therefore
$$\left(\frac{n}{n+1}\right)^n=e^{n\log \left(\frac{n}{n+1}\right)}=e^{-\frac{n}{n+1}\frac{\log \left(1-\frac{1}{n+1}\right)}{-\frac1{n+1}}}$$
with
$$-\frac{n}{n+1}\frac{\log \left(1-\frac{1}{n+1}\right)}{-\frac1{n+1}} \to -1\cdot 1 =-1$$