# For each of the following evaluate the limit or show that the limit does not exist $\lim_{(x, y) \to (0,0)} \frac{\sin(x-y)}{||(x, y)||}$

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.

• 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. – Don Thousand Oct 19 '18 at 2:32
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 $$\lim\limits_{x\rightarrow 0} 0/x^2=0$$.
When $$y=0$$, then $$\lim_{x>0\rightarrow 0}\ \frac{\sin\ x}{x} =1$$ and $$\lim_{x<0\rightarrow 0}\ \frac{\sin\ x}{-x} =-1$$