Limit of partial derivative

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 $(a,b)$.

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

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.
• @S.vanNigtevecht Since $x$ and $\delta y$ are independent of each other, this can be done. Commented Sep 1, 2017 at 9:40