Evaluate $\lim_{x\to \infty} \frac {x^b}{e^{ax}}.$

Question

I would like to determine $$ \lim_{x\to \infty} \frac {x^b}{e^{ax}}. $$

I know for sure that the limit when $x$ approaches infinity is $0.$ But what is the $x$ that's in the power of $e$ is in the power of something different than $1$?

Answer 1

Suppose $a,b,c$ are positive integers. I guess what you mean is to evaluate the limit $\lim_{x\to \infty}\frac{x^c}{e^{ax^b}}$\

Note that this is a limit of the form $\frac{\infty}{\infty}$, we can use the L'hosiptal rule.\

$\displaystyle \lim_{x\to \infty}\frac{x^c}{e^{ax^b}}=\lim_{x\to \infty}\frac{x^{c-b+1}}{abe^{ax^b}}=...$\

Keep doing this, the power in the numerator will become negative and hence the limit will be 0

Answer 2

An idea for you:

$$x^b=e^{b\log x}\implies \frac{x^b}{e^{ax}}=e^{b\log x-ax}$$

Let us now check what happens with $\;b\log x-ax\;$ For example, first case:

$$a=b=0\implies b\log x-ax=0\implies \;\text{the expression and the limit is}\;\;1$$

Second case:

$$a,b>0\implies b\log x-ax\xrightarrow[x\to\infty]{}-\infty\implies \text{ the limit is zero}$$

Continue as above with different cases. Remember: $\;x\to\infty\;$ was faster than $\;\log x\to\infty\;$ when $\;x\to\infty\;$