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Consider the following expression: $$\lim_{(x,y) \to (0,0)} g(x,y)= \begin{cases} \frac{\sin x}{x} y \text{ $\hspace{0.5in}$if $x \neq 0$}\\ y \text{ $\hspace{0.81in}$if $x=0$}\end{cases}$$ I am seeking guidance in regards to a general method for finding limits for piecewise functions such as the one above. Do I take each case individually and find the limit? Since the top function exists only for $x \neq 0$, can the limit exist as $(x,y) \to (0,0)$?

Now say the limit exists for this piecewise function. How would I prove this using the epsilon-delta definition? I am familiar with the process for non-piecewise functions, but this type of question has always confused me. Any general advice (that may assist me in solving more questions like this one) would be appreciated. Thank you!

EDIT: Here is my attempt.

We can directly find that $\lim_{(x,y) \to (0,0)}y=\lim_{y \to 0}y=0$. Now, we will consider the other case. Let $\varepsilon >0$ be arbitrary and take $\delta=\varepsilon$. Then, for all $(x,y)$ with $0<\sqrt{x^2+y^2}<\delta$, we have that $$\left|\frac{\sin x}{x}y\right|\leq |y| \leq \sqrt{x^2+y^2}<\delta=\varepsilon$$ Thus, the limit exists and is $0$.

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Hint:

$\lvert g(x,y) \rvert$ is bounded by a function of $y$ alone.

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    $\begingroup$ So, would I take the limit as $(x,y) \to (0,0)$ for each case, and prove each using the definition? $\endgroup$ – Douglas Fir Oct 4 '15 at 22:32
  • $\begingroup$ Absolutely !... $\endgroup$ – Bernard Oct 4 '15 at 22:35
  • $\begingroup$ Yes, since it is standard that $\lvert\sin x\rvert\le\lvert x\rvert$ (it's a consequence of the mean inequality). $\endgroup$ – Bernard Oct 4 '15 at 22:43
  • $\begingroup$ thank you, it was in my mind $sinx\leq x$ for $0\leq x\leq \frac{\pi}{2}$. $\endgroup$ – R.N Oct 4 '15 at 22:47
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    $\begingroup$ Afortiori, it is true for $x>\frac\pi2$ since $\frac\pi2>1$. $\endgroup$ – Bernard Oct 4 '15 at 22:50

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