Limit of the form 0 times infinity

Question

I'm trying to evaluate $\lim_{n\to\infty} n^3\ln\left(1+\frac{1}{n!}\right)$.

It's $0\cdot\infty$ situation. I know that indeterminate forms can sometimes be evaluated using L'Hopital's rule. I would prefer not to use L'Hopitals rule as there is a factorial and I'm not sure about the next steps.

Are there any methods that can be used to evaluate this kind of limit or any tips on how should I continue?

Answer 1

You can deduce from L'Hopital's Rule that$$\lim_{x\to0}\frac{\log\left(1+x\right)}{x^3}=0,$$which implies that$$\lim_{n\to\infty} n^3\log\left(1+\frac1n\right)=0.$$But$$(\forall n\in\mathbb N):0\leqslant n^3\log\left(1+\frac1{n!}\right)\leqslant n^3\log\left(1+\frac1n\right)$$and so…

Answer 2

You may proceed as follows:

You have

$$n^3\ln (1+\frac 1{n!})= \frac{n^3}{n!}\ln (1+\frac 1{n!})^{n!}\stackrel{n\to\infty}{\longrightarrow}0\cdot 1=0$$

Here, I use $\lim_{m\to \infty}(1+\frac 1m)^m = e$ and $\ln e = 1$.