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If $\lim_{x\to \infty}xf(x^2+1)=2$ then find $$\lim_{x\to 0}\dfrac{2f'(1/x)}{x\sqrt{x}}=?$$

My Try : $$g(x):=xf(x^2+1)\\g'(x)=f(x^2+1)+2xf'(x^2+1)$$ Now what?

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  • $\begingroup$ Set $x^2+1=1/y$ $\endgroup$ – lab bhattacharjee Nov 5 '17 at 18:58
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Call $x^2+1=1/t$. Then $$ \lim_{t\to 0^+}\sqrt{1/t-1}f(1/t)=\lim_{t\to 0^+}\frac{f(1/t)}{(1/t-1)^{-1/2}}=2. $$ Using L'Hopital, if the following limit exists (ratio of derivatives upstairs and downstairs) $$ \lim_{t\to 0^+}\frac{(-1/t^2)f'(1/t)}{\frac{1}{2 \left(\frac{1}{t}-1\right)^{3/2} t^2}}=\lim_{t\to 0^+} -2\left(\frac{1}{t}-1\right)^{3/2}f'(1/t) $$ should also be equal to $2$. But this is equal to $$ \lim_{t\to 0^+} -2\frac{f'(1/t)}{t^{3/2}}\ , $$ since $(1-t)^{3/2}\to 1$. Therefore the result you are after is $-2$.

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Note that, by letting $t=x^2+1$, we have $$2=\lim_{x\to \infty}xf(x^2+1)=\lim_{t\to +\infty}\sqrt{t-1}f(t)=\lim_{t\to +\infty}\sqrt{t}f(t).$$ Hence, if the desired limit exists, then, by letting $t=1/x$, \begin{align} \lim_{x\to 0^+}\frac{2f'(1/x)}{x\sqrt{x}}&= \lim_{t\to+\infty}2t\sqrt{t}f'(t) =\lim_{t+\infty}t^{2}f(t)f'(t)\\ &=-\frac{1}{2}\lim_{t\to +\infty}\frac{(f^2(t))'}{(1/t)'}=-\frac{1}{2}\lim_{t\to +\infty}\frac{f^2(t)}{1/t}\\&=-\frac{1}{2}\left(\lim_{t\to+\infty}\sqrt{t}f(t)\right)^2=-\frac{4}{2}=-2 \end{align} where we used L'Hôpital's rule in the second line.

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