# $\lim_{n\to +\infty}\ e^{\sqrt n } * \left(1 - \frac{1}{\sqrt n}\right)^n$

$$\lim_{n\to +\infty}\ e^{\sqrt n } * \left(1 - \frac{1}{\sqrt n}\right)^n$$

A)0

B)1

C)e

D)${\sqrt e}$

E)$\frac{1}{\sqrt e}$

I tried working on the second part to get it at a better form. In the end I got $e^{- \sqrt n}$. Returning at the beginning with this new form it would be: $$\lim_{n\to +\infty}\ e^{\sqrt n } * e^{- \sqrt n}$$ which would eventually turn into $$\lim_{n\to +\infty}\ e^{0}$$ so the answer would be B) 1 but it's not. The answer is E)$\frac{1}{\sqrt e}$ but I can't figure it out why.

Consider $$A= e^{\sqrt n } * \left(1 - \frac{1}{\sqrt n}\right)^n$$ Take logarithms of both sides $$\log(A)=\sqrt n+n\log\left(1 - \frac{1}{\sqrt n}\right)$$ Now, use the fact that for small values of $x$, $\log(1-x)=-x-\frac{x^2}{2}+O\left(x^3\right)$. Replace $x$ by $\frac{1}{\sqrt n}$ which makes $$\log(A)=\sqrt n+n(-\frac{1}{\sqrt n}-\frac{1}{2 n}+\cdots)$$ Expand and simplify.

I am sure that you can take from here.

HINT:

Use $e^x=\sum_{r=0}\dfrac{x^r}{r!}$ for $e^{\sqrt n}$

and Binomial series for $\left(1-\dfrac1{\sqrt n}\right)^n$

The answer that I worked out is $E)$. For the details, some key points: $1)$: Put $k = \sqrt{n}, x = \dfrac{1}{k}$ then you are led to the use of L'hopitale rule twice for $f(x) = \dfrac{x+\ln(1-x)}{x^2}$ after taking Log.

• But it's not the D) (no joke intended). – user254034 Jul 15 '15 at 8:40
• Yes, you are right, I missed the minus $-$ when taking second derivative. – DeepSea Jul 15 '15 at 8:44