Calculating $\lim_{x \rightarrow +\infty} \frac{e^{ax}}{x^b}$ using de l'Hospital

Question

I am trying to calculate $\lim_{x \rightarrow +\infty} \frac{e^{ax}}{x^b}$ where a,b $>$0

Because $\lim_{x \rightarrow +\infty} x^b=+\infty$ I can use l'Hospitals Rule. So, $\lim_{x \rightarrow +\infty} \frac{e^{ax}}{x^b}=\lim_{x \rightarrow +\infty}\frac{ae^{ax}}{bx^{b-1}}$ but this new limit is not useful. So after some tries I figured out that, writting $\frac{e^{ax}}{x^b}$ as $(\frac{e^{\frac{a}{b}x}}{x})^b$ could be promising

So now I tried to calculate $lim_{x \rightarrow \infty}\frac{e^{\frac{a}{b}x}}{x}=lim_{x \rightarrow \infty}\frac{\frac{a}{b}e^{\frac{a}{b}x}}{1}\rightarrow \infty$ (1)

Now, because of $a^b>a$ for $a>1,b>1$ one can argue that from (1) follows $lim_{x \rightarrow \infty}(\frac{e^{\frac{a}{b}x}}{x})^b=\infty$ for $b>1$

My Question is, how to calculate this limit where $0<b<1$

Answer 1

Use L'Hopital's rule $\lceil b \rceil$ times to get $$\lim_{x\to\infty}C\frac{e^{ax}}{x^{b-\lceil b\rceil}}=0.$$

Answer 2

It goes to infinity for all $a$ and $b$ that are greater than $0$. This is because the exponential grows faster than any polynomial eventually. This holds also for $b\leq 1$.

Answer 3

Alternative approach.

You can compute $\lim_{x\to \infty} \frac{f(x)}{g(x)}$ by (instead) computing $\lim_{x \to \infty} \log\left[\frac{f(x)}{g(x)}\right].$

$$\log\left[\frac{e^{ax}}{e^{b\log x}}\right] = ax - b\log(x).\tag1$$

Ironically, using L'Hopital's rule, $\lim_{x \to \infty} \frac{x}{\log(x)} = \infty.$

Thus, when evaluating the RHS of (1) above, it is immediate that regardless of the values of $a,b$, as long as $a \neq 0$, then $ax$ is the dominating term, and the evaluation will merely depend on the sign (positive or negative) of $a$.