How to find the limit $\lim\limits_{m\to\infty}\frac{m^{m-2}}{(m-1)^{m-2}}$? I am trying to evaluate the limit of this:
$$\lim_{m\rightarrow \infty} \frac{m^{m-2}}{(m-1)^{m-2}}$$
That is just basic calculus I think but I forget those methods for finding the limit. I think the limit is just $e$. Anyone could prove that? Thanks!
Fei
 A: $$\frac{m^{m-2}}{(m-1)^{m-2}}=\left(1+\frac{1}{m-1}\right)^{m-2}\longrightarrow e$$
as $m\rightarrow\infty$.
A: $$\lim_{m\to \infty} \frac{m^{m-2}}{(m-1)^{m-2}}=\lim_{m\to \infty}\left( \frac{m}{m-1}\right)^{m-2}=\lim_{m\to \infty}\left( \frac{m-1+1}{m-1}\right)^{m-2}=$$
$$=\lim_{m\to \infty}\left( 1+\frac{1}{m-1}\right)^{m-1-1}=\lim_{m\to \infty}\left( 1+\frac{1}{m-1}\right)^{m-1}\left( 1+\frac{1}{m-1}\right)^{-1}=$$
$$=\lim_{m\to \infty}\left( 1+\frac{1}{m-1}\right)^{m-1}\lim_{m\to \infty}\left( 1+\frac{1}{m-1}\right)^{-1}=e\cdot 1=e$$
A: You can also use continuity of the exponential function: note that
$$
\ln\left[\left(1+\frac{1}{m-1}\right)^{m-2}\right]=(m-2)\ln\left[1+\frac{1}{m-1}\right]=\frac{\ln\left[1+\frac{1}{m-2}\right]}{\frac{1}{m-2}}.
$$
Both the numerator and denominator tend to 0 as $m\rightarrow\infty$; so, using L'Hopital's rule and the identities
$$
\frac{d}{dm}\ln\left[1+\frac{1}{m-2}\right]=-\frac{1}{(m-1)(m-2)}
$$
and
$$
\frac{d}{dm}\left[\frac{1}{m-2}\right]=-\frac{1}{(m-2)^2},
$$
we find
$$
\lim_{m\rightarrow\infty}\frac{\ln\left[1+\frac{1}{m-2}\right]}{\frac{1}{m-2}}=\lim_{m\rightarrow\infty}\frac{(m-2)^2}{(m-1)(m-2)}=1.
$$
Now, by continuity of the exponential function, if $\ln f(m)\rightarrow 1$ then $f(m)=e^{\ln f(m)}\rightarrow e^1=e$.
A: $$\frac{m^{m-2}}{(m-1)^{m-2}}=\left(\frac{m}{m-1}\right)^{m-2}=\left(\frac{m-1+1}{m-1}\right)^{m-2}=\left(1+\frac{1}{m-1}\right)^{m-2}$$
As $m\to \infty$, $m-1\to \infty$, so you can replace the limit of this with $$\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{m-1}=\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{m}\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{-1}=e\cdot 1=e$$
