Prove $\sin(xy)/x = a$ when $(x,y) \to (0,a)$, using the epsilon-delta definition.

Question

$$|\sin(xy)/x - a| \leq |\sin(xy)/x| + |a| < |y| + |a|.... $$

Then?

Answer 1

Hint: $\frac{\partial \sin(xy)}{\partial x}$ $=y\cos(xy)$, which is $y$ at $x=0$. So for any fixed $y$, the Taylor series of $\sin(x)$ around $x=0$ is $yx + o(x)$.

Finish via L Hopital's rule for $(x,a) \rightarrow (0,a)$. Then conclude that this implies $\sin(xy)/x \rightarrow a$ when $(x,y) \rightarrow (0,a)$. [Let $(x_n,y_n)$ be a sequence whose limit point is $(0,a)$. Use L' Hopital to conclude that for each $\epsilon'$, there is a $\delta'$ s.t. inequality $|\frac{\sin(x_ny_n)}{x_n} - y_n| \leq \epsilon'$ holds as long as both $x_n$ and $y_n$ satisfy $|x_n| \leq \delta'$ and $|y_n-a| \leq \delta'$, and that there is an $l$ s.t. $|x_n| \leq \delta'$ and $|y_n-a| \leq \delta'$ for all $n \ge l$. Furthermore, one can conclude WLOG that $\delta' \le \epsilon'$.

Then use this to conclude that, given $\epsilon$, set $\epsilon' = \frac{\epsilon}{2}$ and use the above to conclude that there exists an $l$ s.t for all $n \ge l$, the inequality $|\frac{\sin(x_ny_n)}{x_n} - a| \leq \delta'+\epsilon' \le 2\epsilon' = \epsilon$ holds if $n$ is at least $l$. Then use this to conclude $\frac{\sin(xy)}{x} \rightarrow a$ when $(x,y) \rightarrow (0,a)$.]