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Let $S(x,y)$ be a postive real value function defined on $(0,\infty)\times(a,b)$ that has both partial derivatives and satisfies $\lim_{x\rightarrow 0} S(x,y)=0$ for every given $y$ in $(a,b)$.

Prove or disprove that $$\lim_{x\rightarrow 0} \frac{\partial S}{\partial y}=0$$ for every $y$ in $(a,b)$.

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    $\begingroup$ it is advisable to include our attempt when we post a question on this site. $\endgroup$ Commented Sep 1, 2017 at 1:15

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We know that $$\frac{\partial S}{\partial y} = \lim_{\delta y\rightarrow0}\left(\frac{S(x, y+\delta y)-S(x,y)}{\delta y}\right)$$ Hence, $$\begin {align} \lim_{x\rightarrow0}\left(\frac{\partial S}{\partial y}\right) & = \lim_{x\rightarrow0}\left(\lim_{\delta y\rightarrow0}\left(\frac{S(x, y+\delta y)-S(x,y)}{\delta y}\right)\right)\\ &=\lim_{\delta y\rightarrow 0}\left(\lim_{x\rightarrow0}\left(\frac{S(x, y+\delta y)-S(x,y)}{\delta y}\right)\right)\\ &=\lim_{\delta y\rightarrow0}\left(\frac{\displaystyle{\lim_{x\rightarrow0}}(S(x,y+\delta y))-\displaystyle{\lim_{x\rightarrow0}}(S(x, y))}{\delta y}\right)\\ &=\lim_{\delta y \rightarrow0}\left(\frac{0-0}{\delta y}\right)\\ &=0 \end{align}$$ Hence the proof is complete.

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  • $\begingroup$ @S.vanNigtevecht Since $x$ and $\delta y$ are independent of each other, this can be done. $\endgroup$
    – Faiq Irfan
    Commented Sep 1, 2017 at 9:40
  • $\begingroup$ I think my case satisfies the conditions to swap limits. $\endgroup$
    – Faiq Irfan
    Commented Sep 1, 2017 at 9:48
  • $\begingroup$ Usually, it requires at least one of the limit to be uniformly convergent to justify the eligibility of interchanging limits. Can you elaborate more on that? $\endgroup$ Commented Sep 1, 2017 at 16:37

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