As in the title here is the limit:

$$ \lim_{n \rightarrow \infty} \left( 1- \left( 1- \frac{c}{n} \right)^{\frac{1}{2}} \right)^{\frac{n}{2}}$$

I tried putting it in a form where I could utilize the fact that $\lim\limits_{n \rightarrow \infty} \left( 1+ \frac{1}{n} \right)^{n} = e$ but was not successful.

EDIT: c is in fact < 0. so the accepted answer is the simpler case, I got a bit confused between my notes and deciding to call the constant c. I don't expect a full response after my mistake I am just editing for completeness.

  • $\begingroup$ Is $c>0$ by any chance? $\endgroup$ – rtybase Sep 15 '17 at 15:35
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    $\begingroup$ Also $\lim\limits_{n \rightarrow \infty}\left(1−\frac{1}{n}\right)^n \ne e$ $\endgroup$ – rtybase Sep 15 '17 at 15:51
  • $\begingroup$ Please say what $c$ is. $\endgroup$ – zhw. Sep 15 '17 at 16:37

$1-\left(1-\frac{c}{n}\right)^{\frac{1}{2}}\rightarrow0$. I hope now it's clear.


Hint. Note that $$ 1-\left( 1- \frac{c}{n} \right)^{1/2} \to 0^+.$$ Moreover, $(0^+)^{+\infty}$ is not an indeterminate form.


I hope Robert and Michael won't mind me elaborating a little bit further to their solutions (which I had in mind too, but I am not so fast at typing)

We need to assume that $c>0$, otherwise things aren't so smooth. From some $n$ onwards, we have $0<1-\left(1-\frac{c}{n}\right)^{\frac{1}{2}}<1$, to be more precise, from $0<c<n$.

Then we have $$1-\left(1-\frac{c}{n+1}\right)^{\frac{1}{2}} < 1-\left(1-\frac{c}{n}\right)^{\frac{1}{2}}$$

Then, from some $n$ onwards, we can assume the existence of an $\alpha$ such that $$0<1-\left(1-\frac{c}{n}\right)^{\frac{1}{2}} < \alpha < 1$$

Then $$0<\left(1-\left(1-\frac{c}{n}\right)^{\frac{1}{2}}\right)^{\frac{n}{2}} < \alpha^{\frac{n}{2}} < 1$$

And then use the squeeze theorem to show the limit is $0$.


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