How to calculate $ \lim\limits_{n\to\infty} e^n\left( 1-\frac{1}{\sqrt n} \right)^n $?

Question

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

How is this limit? And how does it work?

Answer 1

Note that $1-1/\sqrt n > 1/2$ for $n>4.$ For such $n$ the expression is at least $e^n\cdot (1/2)^n = (e/2)^n \to \infty.$

Answer 2

The logarithm of the expression is $$n+n\ln\left(1-\frac1{\sqrt n}\right).$$ Recall that $$\ln(1-x)=-x+O(x^2)$$ for $x\to0$ etc.

Answer 3

If $0 < c < 1$, let $f(n) =e^n \left( 1-\frac{1}{n^{1-c}} \right)^n $.

Then, using $\ln(1-x) = -x+O(x^2)$ for small $x$,

$\begin{array}\\ \ln(f(n)) &=n+n\ln(1-\frac{1}{n^{1-c}})\\ &=n+n(-\frac{1}{n^{1-c}}+O(\frac{1}{n^{2-2c}}))\\ &=n-n^c+O(\frac{1}{n^{1-2c}})\\ \end{array} $

Since $c < 1$, $n^c = o(n)$ so $n-n^c \to \infty$.

If $c > \frac12$, $\frac{1}{n^{1-2c}} =n^{2c-1} =o(n) $ since $2c-1 < 1$.

Therefore $\ln(f(n)) = n-o(n) \to \infty$ so $f(n) \to infty$.

Your question is the case $c = \frac12$.