# limits of $\lim_{(x,y)\to(0,0)}\frac{x\sin{1/x}+y}{x+y}$

How to calculate this limit: $$\lim_{(x,y)\to(0,0)}\frac{x\sin{\frac{1}{x}}+y}{x+y}$$ I have found that $$|\frac{x\sin{\frac{1}{x}}+y}{x+y}|\leq1$$ but I can't conclude.

• Hint: Try taking the limit along lines to the origin (such as $x=0$, $y=0$, or $y=mx$).
– kccu
Apr 29 '17 at 14:47
• Take the line $y=0$. Then $$\lim_{x\to0}\frac{x\sin{\frac{1}{x}}+y}{x+y}=\lim_{x\to0}\sin\left(\frac1x\right)$$ Apr 29 '17 at 14:48
• Then, there is no limit? Apr 29 '17 at 14:49
• @TheoryNombre Nope Apr 29 '17 at 14:50

It doesn't exist:

$$\lim_{x\to0}\lim_{y\to0}\frac{x\sin\frac1x+y}{x+y}=\lim_{x\to0}\frac{x\sin\frac1x}x=\lim_{x\to0}\sin\frac1x$$

If it doesn't exist along one path, it doesn't exist at all.

Take $x=y$.

It becomes

$$\lim_{x\to 0}\frac {\sin (\frac {1}{x})+1 }{2}$$

if $x=\frac {1}{n\pi}$, we find $\frac {1}{2}$

and if $x=\frac {1}{ \frac {\pi}{2}+2n\pi }$, we find $1$.

the limit doesn't exist.

• @TheoryNombre not at all friend of the world. Apr 29 '17 at 15:00