Computing the limit of $\lim_{t\rightarrow0}tf(g(t))$ assuming $g(0)=0$ and $g'(0)>0$

Question

Suppose $f:(0,\infty)\rightarrow\mathbb{R}$ is a continuous function and $g:\mathbb{R}\rightarrow\mathbb{R}$ is a $C^1$ function with $g(0)=0$ and $g'(0)>0$. If the limit $$ \lim_{t\rightarrow0^+} tf(t)=a $$ exists, can we necessarily compute the limit $$ \lim_{t\rightarrow0^+} tf(g(t))? $$ It seems like we can compute it as \begin{align*} \lim_{t\rightarrow0^+} tf(g(t)) &= \lim_{t\rightarrow0^+} tf\left(t\frac{g(t)}{t}\right) \\&= \lim_{t\rightarrow0^+} tf\left(tg'(0)\right) = \frac{1}{g'(0)}\lim_{t\rightarrow0^+}tf(t) = \frac{a}{g'(0)} \end{align*} but I'm not sure the step from line 1 to line 2 is valid. Is it?

Answer 1

You get the correct value of the limit. The same answer is obtained by: \begin{align} \lim_{t\to 0^+}tf(g(t)) &=\lim_{t\to 0^+}\frac{t}{g(t)}g(t)f(g(t))\\ &=\lim_{t\to 0^+}\frac 1{g(t)/t}g(t)f(g(t))\\ &=\frac{a}{g'(0)} \end{align} where $g(t)f(g(t))\to a$ because $g(t)\to 0^+$ as $t\to 0^+$.