2
$\begingroup$

Find $$\lim\limits_{n \to \infty}\left(\frac{\alpha}{x/n + \alpha}\right)^{\beta n}$$ where $\alpha, \beta, x > 0$.

I have a solution here that says that it is $e^{-x\beta / \alpha}$, but it isn't easy for me to see this from the limit definition: we would have $$e^{-x\beta / \alpha} = \lim_{n \to \infty}\left(1+\dfrac{-x\beta/\alpha}{n} \right)^{n}\text{.}$$ Is there a fancy substitution that I'm not seeing that shows that these two forms are equivalent?

Note that this was for a timed qualifying exam, so efficient solutions are preferable.

$\endgroup$
4
$\begingroup$

One may write $$ \left(\frac{\alpha}{x/n +\alpha}\right)^{\beta n}=\left[\left(1+\frac{x/\alpha}{n} \right)^n\right]^{-\beta} $$ then let $n \to \infty$.

$\endgroup$

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.