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

AxiomOfChoice

I'm trying to show the following:

[tex]
\lim_{(x,y) \to (0,0)} \frac{x^2 + \sin^2 y}{x^2 + y^2} = 1.
[/tex]

One can show that

[tex]
\frac{x^2 + \sin^2 y}{x^2 + y^2} \leq 1
[/tex]

for all [itex]x,y[/itex] because [itex]\sin y \leq y[/itex]. So, if you can bound this guy from below by something that goes to 1 as [itex](x,y) \to (0,0)[/itex], 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

[tex]\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))[/tex] where θ is the angle of (x,y). Does that help at all? Edit: I put a mult. sign where the + sign should be! o_O

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.