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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?

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  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh Oct 12 '18 at 3:14
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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 $t\to 0^+$.

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

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