# limit of series using l'Hopitals Rule

Here is an exercise with the solution. I'm curious about how this solution can be found in the proposed way (i.e. using l'Hopital's Rule).

$\lim \limits_{n \to \infty} \left(\frac{(n - 3)}{n}\right)^n$ = $\lim \limits_{n \to \infty} \left(1 + \frac{-3}{n}\right)^n = e^{-3}$ by l-HÔpital's rule.

I see that $e = \lim \limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$ although I don't see how l-HÔpital's rule was applied above.

Grateful for clarification on this

It's well-known that $$e^a=\lim_{n\to\infty}\left(1+\frac an\right)^n.$$ This is equivalent to $$a=\lim_{n\to\infty}n\ln\left(1+\frac an\right).$$ You can apply L'Hospital to $$\lim_{x\to0^+}\frac{\ln(1+ax)}{x}$$ to get this (or just remember the definition of derivative).
The right and the simpler approach to evaluate the limit in your question is as follows $$\left(\frac{n-3}{n}\right)^{n}=\dfrac{1} {\left(\dfrac{n}{n-1}\right)^{n}\left(\dfrac{n-1}{n-2}\right)^{n}\left(\dfrac{n-2}{n-3}\right)^{n}}$$ The first factor in denominator on right side of the above equation can be written as $$\left(1+\frac{1}{n-1}\right)^{n-1}\cdot\frac{n}{n-1}$$ which clearly tends to $e$. Similarly other factors in denominator also tend to $e$ and hence the desired limit is $1/e^{3}=e^{-3}$. The same technique can be used to prove that if $x$ is rational then $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=e^{x}$$