limit of $\frac{x^{\frac{1}{2}}}{(\frac{1}{2})^x}$

Question

I have to compute the following limit: $$\lim_{x\to +\infty}\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)^x},$$ for which we have the indetermination $\frac{+\infty}{0}$.

I know that the final answer is $+\infty$. However, I am not sure if the way I am solving it is correct $$\lim_{x\to +\infty}\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)^x}=\lim_{x\to +\infty} 2^xx^{\frac{1}{2}}=\lim_{x\to +\infty} 2^x . \lim_{x\to +\infty} x^{\frac{1}{2}} = +\infty.+\infty=+\infty,$$ because I know there are some issues when computing $\lim_{x\to \infty} 1^x$, therefore I do not know if doing $\lim_{x\to +\infty}\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)^x}=\lim_{x\to +\infty} 2^xx^{\frac{1}{2}}$ is correct?

Thank you very much!

Answer 1

$\frac{\infty}{0}$ is not an indeterminate form.

Answer 2

I think it's correct. $\left (\frac{1}{2} \right )^{x} = \frac{1}{2^{x}}$, and for each $x$, we have $$\frac{x^{\frac{1}{2}}}{\frac{1}{2^{x}}} = x^{\frac{1}{2}}2^{x}$$

and since these are the same for each $x$, they have the same limit, i.e.,

$$\lim \limits_{x \to \infty} \frac{x^{\frac{1}{2}}}{\frac{1}{2^{x}}} = \lim \limits_{x \to \infty} x^{\frac{1}{2}}2^{x}.$$

It's kind of like saying: if $f(x) = g(x)$ for every $x$ (i.e., $f(x)$ and $g(x)$ are the same function) then $\lim \limits_{x \to \infty} f(x) = \lim \limits_{x \to \infty} g(x)$. It's obvious that this is true if $f(x)$ is the same function as $g(x)$.