limit of $(1-\frac{x}{\sqrt{n}})^{-n}$ as n goes to infinity

Question

I am looking to find the limit of this
$$lim_{n\rightarrow\infty}(1-\frac{x}{\sqrt{n}})^{-n}$$

I know that we have $$lim_{n\rightarrow\infty}(1+\frac{x}{n})^n=e^x$$ can anyone help?

Answer 1

You can write:

\begin{align*} \lim_{n\to+\infty}\left(1-\frac{x}{\sqrt{n}}\right)^{-n}&= \lim_{n\to+\infty}\left(1+\frac{1}{-\frac{\sqrt{n}}{x}}\right)^{-n}\\ &=\lim_{n\to+\infty}\left[\left(1+\frac{1}{-\frac{\sqrt{n}}{x}}\right)^{-\frac{\sqrt{n}}{x}}\right]^{\frac{x}{\sqrt{n}}n}.\\ \end{align*} The exponent $\dfrac{xn}{\sqrt{n}}=x\sqrt{n}$ goes to $+\infty$ if $x>0$, so the limit is infinite, and goes to $-\infty$ if $x<0$: in this case the limit goes to $0$. If $x=0$ then the limit is $1$.

Answer 2

Note that

$$\left( 1-\frac{x}{\sqrt{n}}\right)^{-n}=e^{ -n\log \left( 1-\frac{x}{\sqrt{n}}\right)}$$

and since by standard limits as $x\to 0, \,\frac{\log(1+x)}{x}\to 1$

$$-n\log \left( 1-\frac{x}{\sqrt{n}}\right)=\frac{nx}{\sqrt n}\frac{\log \left( 1- \frac{x}{\sqrt{n}}\right)}{-\frac{x}{\sqrt{n}}}\to \begin{cases}+\infty \,\text{for}\,x>0\\\\-\infty\,\text{for}\,x<0\end{cases}$$

therefore

• for $x>0 \implies\left( 1-\frac{x}{\sqrt{n}}\right)^{-n}\to \infty$

• for $x<0 \implies\left( 1-\frac{x}{\sqrt{n}}\right)^{-n}\to 0$