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For each of the following evaluate the limit or show that the limit does not exist

$\lim_{(x, y) \to (0,0)} \frac{\sin(x-y)}{\|(x, y)\|}$

Solution:

$=\lim_{(x, y) \to (0,0)} \frac{\sin(x-y)}{\sqrt{x^2 + y^2}}$

If we let $y = x$, then

$=\lim_{x \to 0)} \frac{\sin(0)}{\sqrt{2x^2}} = 0$ [Wouldn't this be indeterminate $0/0$?]

If we let $y = -x$, then

$\lim_{x \to 0^+} \frac{\sin(2x)}{\sqrt{2x^2}} = \lim_{x \to 0^+} \frac{\sqrt{2}\sin(2x)}{2x} = \sqrt{2}$

and since they are different values we know the limit does not exist.

Could someone please explain

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    $\begingroup$ Try setting $y=0$, and then $x=0$, and take the limits as the other variable goes to 0. You'll get two different limits. $\endgroup$ – Don Thousand Oct 19 '18 at 2:32
  • $\begingroup$ The limit is unique, then if you have found two different values for the limit, it is not be exist. $\endgroup$ – Nosrati Oct 19 '18 at 2:36
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It wouldn't be indeterminate, remember that in a limit does not matter the value of the object in the point, only matters what happens as we approach it. So $\lim\limits_{x\rightarrow 0} 0/x^2=0$.

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When $y=0$, then $$\lim_{x>0\rightarrow 0}\ \frac{\sin\ x}{x} =1 $$ and $$\lim_{x<0\rightarrow 0}\ \frac{\sin\ x}{-x} =-1 $$

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