evaluate $\lim_{(x,y) \to (0,0)} \frac{\sin(x)\sin(3y)}{2xy}$

Question

I'm studying calculus of multiple variables and I can solve limits of the form $x^2+y^2$, but I need to solve this.

$$\lim_{(x,y) \to (0,0)} \frac{\sin(x)\sin(3y)}{2xy}$$

Could you give me an idea of how to solve that limit?

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Answer 1

As written, the quotient isn't defined if $x = 0$ or $y = 0$. Excluding this subset, you can use that $\lim_{t\to 0}\frac{\sin t}{t} = 1$.

Answer 2

As was written in the comments, you could use the fact that limits are multiplicative to conclude $$\begin{split} \lim_{(x,y)\to(0,0)} \frac{\sin x \sin 3y}{2xy} &= \frac 1 2 \cdot \lim_{(x,y)\to(0,0)} \frac{\sin x}{x} \cdot \lim_{(x,y)\to(0,0)} \frac{\sin 3y}{y} \\ &= \frac 3 2 \cdot \lim_{x \to 0}\frac{\sin x}{x} \cdot \lim_{y \to 0} \frac{\sin 3y}{3y} \\ &= \frac 3 2 \cdot 1 \cdot 1 = \frac 3 2. \end{split}$$ Alternatively, you could solve it directly by seeing that, as $(x,y) \to (0,0)$, we have the first-order expansion $\sin(x) \simeq x$ and $\sin 3y \simeq 3y$, so that $$\lim_{(x,y)\to(0,0)} \frac{\sin x \sin 3y}{2xy} = \lim_{(x,y)\to(0,0)} \frac{3xy}{2xy} = \frac 3 2. $$