How to solve the limit $\lim_{n\to\infty}\frac{n^n}{e^n}$.

Question

Please help me to evaluate this limit. $$\lim_{n\to\infty}\frac{n^n}{e^n}$$ I am totally confused about this. I tried with $L'Hospital$ but failed.

Answer 1

$\frac{n^n}{e^n}=(\frac{n}{e})^n$ and for $n\geq3$ the base is greater than $1$. Conclusion?

Answer 2

This diverges pretty violently and can be shown to do so in many ways.

Here is one of them:

If $n > 2e$ then $\dfrac{(n)^n}{e^n} \gt \dfrac{(2e)^n}{e^n} =2^n \to \infty $.

Answer 3

$$\lim _{ n\rightarrow \infty }{ \frac { n^ n }{ e^ n } } =\lim _{ n\rightarrow \infty }{ (\frac{n}{e})^n } $$

Apply exponent rule: $a^n=e^{\ln(a^n)}=e^{n.\ln(a)}$

In this case we get: $$(\frac{n}{e})^n=e^{n\ln(\frac{n}{e})}$$

$$=\lim _{ n\rightarrow \infty }(e^{n\ln(\frac{n}{e})})$$

Apply the Limit Chain Rule: if $\lim _{ u\rightarrow b }f(u)=L,$ and $\lim _{ x\rightarrow a }g(x)=b,$ and $f(x)$ is continuous at $x=b$. Then: $\lim _{ x\rightarrow a }f(g(x))=L$

In this problem, $g(n)=n\ln(\frac{n}{e}),f(u)=e^u$

$$\lim _{ n\rightarrow \infty }(n\ln(\frac{n}{e}))=\infty$$ After simplifying, $$\lim _{ n\rightarrow \infty }(e^n)=\infty$$