Evaluate $\lim\limits_{\alpha \to \infty} e^{-\frac{t}{\sqrt{\alpha}}}(1-\frac{t}{\sqrt{\alpha}})^{-\alpha}$ [duplicate]

How does one show

$$\lim_{\alpha \to \infty} e^{-t\sqrt{\alpha}}\left(1-\frac{t}{\sqrt{\alpha}}\right)^{-\alpha} = e^{t^2 / 2}?$$

Not homework, this is from this proof that the gamma distribution has a limiting distribution of the standard normal as $\alpha \to \infty$. It suggests using numerical techniques to find the limit of the above, but I would like to know if there is a good way to solve this manually. I tried L'Hopital's rule but got $e^t$, which is obviously incorrect.

marked as duplicate by Guy Fsone, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 11 '17 at 6:07

• +1 this looks like a limit worth scratching your head about! – Zach466920 Jun 24 '15 at 23:50
• Duplicate of this and this. – Lucian Jun 25 '15 at 0:40

$$\exp\left(\lim_{\alpha\to\infty}-t\sqrt\alpha+\alpha\log\left(1+{-t\over\sqrt\alpha}\right)\right).$$

Now let $\beta=\alpha^{-1}$.

$$\exp\left(\lim_{\beta\to 0^+}-t\beta^{-1/2}-\beta^{-1}\log\left(1+(-t\sqrt\beta)\right)\right).$$

Now here we can use the Maclaurin series for $\log(1+x)$ since we're going to $0$. We get

$$\exp\left(\lim_{\beta\to 0^+}-t\sqrt\beta^{-1/2}-\beta^{-1}(-t\sqrt\beta-{t^2\beta\over 2}+O(\beta^{3/2})\right).$$

The only term which doesn't vanish is ${t^2\over 2}$ so we recover the result.

• Slick! I wish I came to this myself but thank you. – mathjacks Jun 25 '15 at 1:27
• @mathjacks glad to help. It was a fun problem, thanks for posting! – Adam Hughes Jun 25 '15 at 1:30