$$\lim_{x\to ∞} \frac {x^{1000000}} {e^x}$$
could anyone please provide some hits with what result I will end up?
After all applyings of L'Hospital rule, I will get $\frac {n} {e^x}$, where $n$ is large number before I got out of the $x$ powers. So, will it be the limit $0$ then? Since the infinity is nothing I have $\frac {n} {0}.$ Or will it be just the $\infty$?
Best Answer
The limit will be 0. Another way to see this:
Note that $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots + \frac{x^{1000001}}{1000001!} + \cdots > \frac{x^{1000001}}{1000001!}$$ $$0<\frac{x^{1000000}}{e^x} < \frac{1000001!}{x}$$
$$\lim_{x \to \infty} \frac{1000001!}{x} = 1000001! \lim_{x \to \infty} \frac1x = 0$$
The Squeeze Theorem will give us this result.