2
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Hint: Try taking the limit along lines to the origin (such as $x=0$, $y=0$, or $y=mx$). $\endgroup$
    – kccu
    Apr 29 '17 at 14:47
  • 2
    $\begingroup$ Take the line $y=0$. Then $$\lim_{x\to0}\frac{x\sin{\frac{1}{x}}+y}{x+y}=\lim_{x\to0}\sin\left(\frac1x\right)$$ $\endgroup$
    – John Doe
    Apr 29 '17 at 14:48
  • $\begingroup$ Then, there is no limit? $\endgroup$ Apr 29 '17 at 14:49
  • $\begingroup$ @TheoryNombre Nope $\endgroup$
    – John Doe
    Apr 29 '17 at 14:50
4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @TheoryNombre not at all friend of the world. $\endgroup$ Apr 29 '17 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.