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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.

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    $\begingroup$ 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 $\sin(y/x) = \sin(\pi/2) = 1$. $\endgroup$ – Bungo Dec 17 '18 at 7:52
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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 !

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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.

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