Solving limits of form $\infty \times \infty$

Question

I need help with solving $f(x) \lim_{x \to \infty} e^x\times x^3$ I know that $\to \infty$ is the answer but I don't know how to get there. Using since $e^x \to \infty$ and $x^3 \to \infty$ I get $\infty \times \infty$ which isn't allowed. I guess this needs to be rewritten somehow before taking the limit, but how?

Answer 1

This is one place where going back to the definitions will help. Try to find, for all $M>0$, some $x_0>0$ such that $x>x_0 \implies e^x x^3>M$.

Use the facts you've indicated in your question.

Answer 2

Hint: $x^3\ge 1$ for $x\ge 1,$ so since $e^x>0$ for all $x,$ then $e^xx^3\ge e^x$ for $x\ge 1.$ Now, use this to show that for all $M$ there is some $N>0$ such that $e^xx^3\ge M$ whenever $x\ge N$.

Added Hint: Find $N'>0$ such that $e^x\geq M$ whenever $x\geq N'$, then put $N=\max\{1,N'\}.$ Now show that $x^3e^x\geq M$ whenever $x\geq N.$

Answer 3

Let $\tilde M := \max(M,1)$ (just technical) Chose $N = \log(\tilde M) + 1$ then for $x > N$ we have $$e^x x^3 > e^{\underbrace{\log\tilde M + 1}_{\geq \log M}} (\underbrace{\log\tilde M}_{\geq \log 1 = 0} + 1)^3 \geq e^{\log M} = M$$ So given any $M$, we can find $N$ such that the expression exceeds $M$ for all $x > N$. Thus $\to\infty$.