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.
Asked
Active
Viewed 37 times
2
-
1Hint: 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
-
2Take the line $y=0$. Then $$\lim_{x\to0}\frac{x\sin{\frac{1}{x}}+y}{x+y}=\lim_{x\to0}\sin\left(\frac1x\right)$$ – John Doe Apr 29 '17 at 14:48
-
Then, there is no limit? – Theory Nombre Apr 29 '17 at 14:49
-
@TheoryNombre Nope – John Doe Apr 29 '17 at 14:50
2 Answers
4
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.
Simply Beautiful Art
- 74,685
2
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.
hamam_Abdallah
- 62,951