Question

Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine another approach path. Approach (0, 0) along the curve y = x2. When x is positive, we have lim (x, y) → (0, 0) xy x2 + y2 =

Answer 1

Answer:

This approach to (0,0) also gives the value 0

Step-by-step explanation:

Probably, you are trying to decide whether this limit exists or not. If you approach through the parabola y=x², you get

$\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{\sqrt{x^2+y^2}}=\lim_{(x,x^2)\rightarrow (0,0)}\frac{xx^2}{\sqrt{x^2+(x^2)^2}}=\lim_{x\rightarrow 0}\frac{x^3}{|x|\sqrt{1+x^2}}=0$

It does not matter if x>0 or x<0, the |x| on the denominator will cancel out with an x on the numerator, and you will get the term x²/(√(1+x²) which tends to 0.

If you want to prove that the limit doesn't exist, you have to approach through another curve and get a value different from zero.

However, in this case, the limit exists and its equal to zero. One way of doing this is to change to polar coordinates and doing a calculation similar to this one. Polar coordinates x=rcosФ, y=rsinФ work because the limit will only depend on r, no matter the approach curve.