# Finding the limit by doing the natural log of the numerator and denominator

$$\lim_{x\to\infty} \frac{e^\sqrt{x}}{x^{a}} = \infty$$

To do this I did the natural log of the numerator and denominator and then did l'Hospital's rule.

My question is: it allowed to do the natural log of the numerator and denominator in order to put it in a clear form before evaluating the limit?

Cause I know $6/3=2$, but $\ln\left(6\right)/\ln\left(3\right)$ is not. If it is not allowed what is a way to concretely evaluate this limit.

$$\frac{e^{2x}}{e^x}$$ is a counterexample. After taking the natural logarithm, we get $2$, so the limit is $2$. The original limit is $\infty$.
You may rather write, as $x \to +\infty$, $$\ln \left(\frac{e^\sqrt{x}}{x^a} \right)=\ln \left(e^\sqrt{x}\right)-\ln \left(x^a\right)=\sqrt{x}-a \times \ln x=\sqrt{x}\left(1-a\:\frac{\ln x}{\sqrt{x}} \right)\to +\infty\times 1$$ thus $$\frac{e^\sqrt{x}}{x^a} \to +\infty.$$