# To find the limit of a given function

Find the limit of $$\lim_{(x,y)\to (0,0)}\sin{\frac{y}{x}}$$

I found out that the function tends to $$0$$ as $$\left|\sin{\frac{y}{x}}-0\right| <1$$, but I am not sure whether the method is correct or not.

• While it's true that $|\sin(y/x) - 0| \leq 1$, no tighter bound applies as $(x,y) \to (0,0)$. E.g. if we let $(x,y)$ approach $(0,0)$ along the line $y = (\pi/2)x$, then we have $\sin(y/x) = \sin(\pi/2) = 1$. – Bungo Dec 17 '18 at 7:52

Let $$f(x,y):= \sin (y/x)$$ for $$x \ne 0$$.

Then we have $$f(0,x)=0 \to 0$$ as $$x \to 0$$ and $$f(x,x)= \sin(1) \to \sin(1)$$ as $$x \to 0$$.

Hence, $$\lim_{(x,y) \to (0,0)}f(x,y)$$ does not exist !

We have that

• $$x=y=t \to 0$$

$$\sin{\frac{y}{x}}=\sin 1 \to \sin 1$$

• $$x=t,\,y=t^2,\, t\to 0$$

$$\sin{\frac{y}{x}}=\sin t \to 0$$

therefore the limit doesn't exist.