Let $f, g$ be defined on $(a,\infty)$ and $\lim_{x \to \infty}f(x)=L$ and $\lim_{x \to \infty}g(x)=\infty$, then $\lim_{x \to \infty}f(g(x)) = L$

Question

If $\lim_{x \to \infty} g = \infty$, then for $M>0$, there exists $d_{1} >0$ such that if $x>d_{1}$, then $g(x) > M$. If $\lim_{x \to \infty} f = L$, then given $\epsilon>0$, there exists a $d_{2} >0$ such that if $x>d_{2}$, then $|f(x) - L|<\epsilon$. If we let $M=d_{2}$, then if $x>d_{1}$, we have $g(x)>M=d_{2}$, so then $|f(g(x))-L| < \epsilon$. Hence, $\lim_{x \to \infty} f(g(x)) = L$. Does this look right?