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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

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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).

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L'Hospital's Rule does not apply to sequences, but rather to functions of a continuous real variable and that too under some precise hypotheses (which most users of this rule don't verify as seen in many answers on this site). There is a counterpart to L'Hospital's Rule which goes by the name of Cesaro-Stolz Theorem which does apply to sequences but this does not appear to be useful in the evaluation of limit in question.

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 $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=e^{x}$$

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