Show that f is continous at $(0,0)$ where $ f(x,y)= y\sin(\frac{1}{x})$ if $x \neq 0$ else $0$

Question

Show that f is continous at $(0,0)$ where $ f(x,y)= y\sin(\frac{1}{x})$ if $x \neq 0$ else $0$

$f(0,0)=0$

$0\leq |\sin(\frac{1}{x})|\leq 1$ $\Rightarrow$ $0 \leq |y\sin(\frac{1}{x})| \leq |y|$

now we have $y \to 0$ because$(x,y) \to (0,0)$ therefore by squeeze theorem $\lim_{(x,y) \to (0,0)} |y\sin(\frac{1}{x})| = 0 $ $\Rightarrow$ $\lim_{(x,y) \to (0,0)} y\sin(\frac{1}{x}) = 0 $

function value and limit is equal so continuous...

I am new to multivariable calculus... I wanted to ask if I have done this rightly particularly have I applied the squeeze theorem correctly?

Answer 1

For every $\epsilon >0$ there is a $\delta >0$

such that $(x^2+y^2)^{1/2} \lt \delta$ implies

$|y\sin(1/x)| \lt \epsilon.$

Choose $\delta = \epsilon$.

$|y\sin(1/x)| \le |y| \le (x^2+y^2)^{1/2} \lt \delta =\epsilon.$

Your solution is fine, above a little more formal.