9
$\begingroup$

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

$\endgroup$
  • $\begingroup$ If an answerer answered your question, please mark the answer as accepted. Also, for future reference, please avoid signatures in your questions; Stack Overflow sites are nice enough to sign your content for you. $\endgroup$ – Carter Pape Jul 4 '13 at 4:17
5
$\begingroup$

$$\lim_{m\to \infty} \frac{m^{m-2}}{(m-1)^{m-2}}=\lim_{m\to \infty}\left( \frac{m}{m-1}\right)^{m-2}=\lim_{m\to \infty}\left( \frac{m-1+1}{m-1}\right)^{m-2}=$$ $$=\lim_{m\to \infty}\left( 1+\frac{1}{m-1}\right)^{m-1-1}=\lim_{m\to \infty}\left( 1+\frac{1}{m-1}\right)^{m-1}\left( 1+\frac{1}{m-1}\right)^{-1}=$$ $$=\lim_{m\to \infty}\left( 1+\frac{1}{m-1}\right)^{m-1}\lim_{m\to \infty}\left( 1+\frac{1}{m-1}\right)^{-1}=e\cdot 1=e$$

$\endgroup$
18
$\begingroup$

$$\frac{m^{m-2}}{(m-1)^{m-2}}=\left(\frac{m}{m-1}\right)^{m-2}=\left(\frac{m-1+1}{m-1}\right)^{m-2}=\left(1+\frac{1}{m-1}\right)^{m-2}$$

As $m\to \infty$, $m-1\to \infty$, so you can replace the limit of this with $$\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{m-1}=\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{m}\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{-1}=e\cdot 1=e$$

$\endgroup$
  • $\begingroup$ You are right, Thanks a lot for the answer!! $\endgroup$ – Pew Jul 3 '13 at 21:35
6
$\begingroup$

$$\frac{m^{m-2}}{(m-1)^{m-2}}=\left(1+\frac{1}{m-1}\right)^{m-2}\longrightarrow e$$ as $m\rightarrow\infty$.

$\endgroup$
5
$\begingroup$

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 $f(m)=e^{\ln f(m)}\rightarrow e^1=e$.

$\endgroup$
  • $\begingroup$ I like this answer a lot, nice work. +1 $\endgroup$ – Alex Wertheim Jul 3 '13 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.