find limit if it exists of function of two variables (x^2 * sin^2y) / (x^2 + 2y^2)

**sgcb** · September 17th 2012, 04:35 PM

$\lim_{(x,y)\rightarrow{(0,0)}}\frac{x^2sin^2(y)}{x ^2+2y^2}$

wolframAlpha claims the limit does not exist, but I'm unable to show how it doesn't. Evaluating the limit on the generalized line y=mx results in

$\lim_{x\rightarrow{0}}\frac{sin^2(mx)}{1+2m^2} = 0$

evaluating along the curve y=x^2 gives

$\lim_{x\rightarrow{0}}\frac{sin^2(x^2)}{1+2x^2} = 0$

AFAICT, the limit /is/ 0. What am I missing?

**TheEmptySet** · September 17th 2012, 04:58 PM

Originally Posted by sgcb

$\lim_{(x,y)\rightarrow{(0,0)}}\frac{x^2sin^2(y)}{x ^2+2y^2}$

wolframAlpha claims the limit does not exist, but I'm unable to show how it doesn't. Evaluating the limit on the generalized line y=mx results in

$\lim_{x\rightarrow{0}}\frac{sin^2(mx)}{1+2m^2} = 0$

evaluating along the curve y=x^2 gives

$\lim_{x\rightarrow{0}}\frac{sin^2(x^2)}{1+2x^2} = 0$

AFAICT, the limit /is/ 0. What am I missing?

I would say that I disagree. If you switch to polar coordinates we get

$\lim_{r \to 0}\frac{r^2\cos^2(\theta)\sin^2(r \sin(\theta))}{r^2+r^2\sin^2(\theta)}=\frac{\cos^2 (\theta)\sin^2(r \sin(\theta))}{1+\sin^2(\theta)}$

Using the above we can squeeze this using
$0 \le \frac{\cos^2(\theta)\sin^2(r \sin(\theta))}{1+\sin^2(\theta)} \le \frac{\sin^2(r \sin(\theta))}{2}$

Now by the squeeze theorem the limit must be zero.

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