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Supose that $f:(0,\infty) \times (0,\infty) \to \mathbb{R}$ is a $C^2((0,\infty) \times (0,\infty))$ function, and $$\lim_{t\to +\infty}f(t,s)=g(s)$$, can we conclude that $$\lim_{t\to +\infty}\frac{d}{ds}f(t,s)=g'(s)?$$ I am trying use the Mean Value Theorem but i cant any argument.

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Let $f:(0,\infty) \times (0,\infty) \to \mathbb{R}$ be defined as $f(t,s)=\frac{\sin(t^2s)}{t}$. Then $$ \lim_{t\to \infty}f(t,s)=0 $$ but $$ \lim_{t\to \infty}f_s(t,s)= \lim_{t\to \infty} t \cos(t^2s) $$ does not exist.

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  • $\begingroup$ and supposing that $\lim_{t\to \infty} f_s(t,s)$ exist? $\endgroup$ Oct 14 at 0:03
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    $\begingroup$ My feeling is still "no", but I don't have a counterexample up to now. In many examples it is true, so there should be a setting where it is true. $\endgroup$
    – Gerd
    Oct 14 at 7:04

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