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

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?

• You can't replace $g(t) /t$ with $g'(0)$. The right approach is the one in the answer by Fabio Lucchini. The same answer shows that continuity of $f$ is not needed and further we don't need $g\in C^{1}$. Just $g(0)=0,g'(0)>0$ is sufficient. – Paramanand Singh Oct 12 '18 at 3:14

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 $$t\to 0^+$$.
• +1 for your approach, but the asker's approach is not correct where $g(t) /t$ is replaced by its limit $g'(0)$. – Paramanand Singh Oct 12 '18 at 3:21