Multivariable proof of $lim_{(x,y)\to(a,b)} \frac{sin(f(x,y))}{f(x,y)}=1$

Question

I know the typical result when it's the limit in one variable, but I can't find a multivariable epsilon-delta proof to the following problem:

Let $f:B_r(a,b) \rightarrow \Bbb R$ such that $f$ is defined over $B_r(a,b)-\{(a,b)\}$ and $lim_{(x,y)\to(a,b)} f(x,y) = 0$. Prove that $$lim_{(x,y)\to(a,b)} \frac{sin(f(x,y))}{f(x,y)}=1$$

$B_r(a,b)$ is the ball with radius $r$ and center $(a,b)$

I basically don't know how to take the known limit to an extra dimension. Where should I start? Also if you know of a book where I can find the proof or another website where it's explained it'll be helpful. Thanks a lot.

Answer 1

Let $\epsilon>0$ we know that $\lim_{z \longrightarrow 0}\frac{sin(z)}{z}=1$ thus $\exists \delta_1$ such that $|z|< \delta_1 \Longrightarrow |\frac{sin(z)}{z}-1|< \epsilon$

But $\lim_{x \longrightarrow (a,b)}f(x,y)=0$ and $\delta_1>0$ thus for $\delta_1>0$, $\exists \delta>0$ such that $||x-(a,b)||< \delta \Longrightarrow |f(x,y)|<\delta_1$

If we set $z=f(x,y)$, then we found $\delta>0$ such that $||x-(a,b)||< \delta \Longrightarrow |f(x,y)| <\delta_1 \Longrightarrow |\frac{\sin{f(x,y)}}{f(x,y)}-1|<\epsilon$