A frustrating limit of a function f(x,y)

[AxiomOfChoice]

I'm trying to show the following:

$$\lim_{(x,y) \to (0,0)} \frac{x^2 + \sin^2 y}{x^2 + y^2} = 1.$$

One can show that

$$\frac{x^2 + \sin^2 y}{x^2 + y^2} \leq 1$$

for all $x,y$ because $\sin y \leq y$. So, if you can bound this guy from below by something that goes to 1 as $(x,y) \to (0,0)$, you should be in business by the Sandwich Theorem. But I have so far been unable to do that! Does anyone have any suggestions as to how to proceed?

 

[jbunniii]

Maybe try this inequality, valid for $|y| \leq \pi$. Plot the two functions to see why it's true.
$$\left|\frac{\sin(y)}{y}\right| \geq \left|1 - \frac{1}{\pi}y\right|$$
Squaring and rearranging gives
$$\sin^2(y) \geq y^2 - \frac{2}{\pi}y^3 + \frac{1}{\pi^2}y^4$$
which looks promising because the $y^3$ and $y^4$ terms go to 0 faster than the $y^2$ term.

 

[joeblow]

$$\underset{(x,y)\rightarrow (0,0)}{\lim}=\frac{x^2\cdot \sin^2 y}{x^2+y^2}=\cos^2 \theta \sin ^2 (r\sin \theta))$$ where θ is the angle of (x,y). Does that help at all? Edit: I put a mult. sign where the + sign should be!

 

[jbunniii]

Here's a more direct way:
$$\begin{align}
\left|\frac{x^2 + \sin^2(y)}{x^2 + y^2} - 1\right| &=
\left|\frac{x^2 + \sin^2(y)}{x^2 + y^2} - \frac{x^2 + y^2}{x^2 + y^2}\right| \\
&= \left|\frac{\sin^2(y) - y^2}{x^2 + y^2}\right| \\
\end{align}$$
The goal is to show that the last expression is arbitrarily small as $(x,y) \rightarrow (0,0)$. If $y = 0$ then the expression equals zero. If $y \neq 0$, then
$$\frac{1}{x^2 + y^2} \leq \frac{1}{y^2}$$
and the result follows easily.