Multivariable limits: $f(x,y)=\frac{\sin(x y)}{x}$

Question

$f(x,y)=\frac{\sin(x y)}{x}$ if $x\neq0$

$f(0,y) = 0$

Find the limit at $(0,0)$ if it exists, or prove the limit doesn’t exist.

Answer 1

It is easy by Tayler series solving. Now proof by limit of two variables $ \lim_{y \rightarrow 0} \lim_{x \rightarrow 0} \frac{sin(xy)} {x} =\lim_{x \rightarrow 0} \lim_{y \rightarrow 0} \frac{sin(xy)} {x} =0 $

Answer 2

We have to consider $x\neq0$ and look at how does the mapping behaves in this case. We have to approach the point $(0,0)$. We have two different cases.

If $y=0$ then the mapping is identically $0$, thus the limit is $0$ as $x\to0$.

Otherwise $y\neq0$; in this case we write $$ \lim_{x\to0}\frac{\sin xy}{x} =y\cdot\lim_{x\to0}\frac{\sin xy}{xy}=y $$ which goes to $0$ as $y\to0$.

So the limit exists and it is $0$. Notice in particular the mapping is continuous on the line $x=0$.