Multivariate limit $\lim_{(x,y)\rightarrow (0,0)} \frac{x^{2}\sin^2y}{x^{2}+6y^{2}}$ So I have solved for the $x$-axis and have gotten $0$ and along the $y$-axis and gotten $\dfrac 16$. However Webassign doesn't seem to like the fact that the limit does not exist. Which leads me to presume the limit does exist. Can someone give me a hint on where to proceed from here. Thanks!
$$\lim_{(x,y)\rightarrow (0,0)} \frac{x^{2}\sin^{2}(y)}{x^{2}+6y^{2}}$$
 A: Put $x=r\cos(t)$ and $y=r\sin(t)$.
After noting that $\sin(y)\sim y$ for $y\to 0$ we have:
$$\frac{x^2\sin^2(y)}{x^2+6y^2}\sim r^2\frac{\cos^2(t)\sin^2(t)}{1+5\sin^2(t)}\leq r^2.$$
This proves that the limit there exists and its value is $0$.
A: Since $0\le \frac{x^{2}}{x^{2}+6y^{2}}\le 1,$ $$ 0\le \frac{x^{2}\sin^{2}(y)}{x^{2}+6y^{2}}\le \sin^{2}(y) \forall (x,y)\neq (0,0) $$
By the squeeze theorem, we get
$$\lim_{(x,y)\rightarrow (0,0)} \frac{x^{2}\sin^{2}(y)}{x^{2}+6y^{2}}=0.$$
A: Note that $\sin y \sim y$ for $y \to 0$.
We have:
$\mathop {\lim }\limits_{(x,y) \to (0,0)} \dfrac{{{x^2}{{\sin }^2}y}}{{{x^2} + 6{y^2}}} = \mathop {\lim }\limits_{(x,y) \to (0,0)} \dfrac{{{x^2}{y^2}}}{{{x^2} + 6{y^2}}}$
Applying the AM–GM inequality, we get:
$\left| {\dfrac{{{x^2}{y^2}}}{{{x^2} + 6{y^2}}}} \right| = \left| {xy} \right|.\left| {\dfrac{{xy}}{{{x^2} + 6{y^2}}}} \right| \leqslant \left| {xy} \right|.\left| {\dfrac{{\frac{{{x^2} + 6{y^2}}}{{2\sqrt 6 }}}}{{{x^2} + 6{y^2}}}} \right| = \dfrac{{\left| {xy} \right|}}{{2\sqrt 6 }} \to 0$ when $(x,y)\to0$.
A: Assuming $x \neq 0$, or in other words we do not approach from the $y$ axis then,
$$|\frac{x^{2}\sin^{2}(y)}{x^{2}+6y^{2}}| \leq | \frac{x^{2}\sin^{2}(y)}{x^{2}}|=\sin^2 (y)$$
But $\sin^2 (y) \to 0$ as $y \to 0$. So our limit goes to zero by squeeze theorem so long as we do not approach along the $y$ axis. If you check the $y$ axis case you'll see the limit is also $0$ along that path.
