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

by AxiomOfChoice
Tags: frustrating, function, limit
 P: 529 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?
 Sci Advisor HW Helper PF Gold P: 3,289 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.
 P: 71 $$\underset{(x,y)\rightarrow (0,0)}{\lim}=\frac{x^2\cdot \sin^2 y}{x^2+y^2}=\cos^2 \theta \sin ^2 (r\sin \theta))$$ where θ is the angle of (x,y). Does that help at all? Edit: I put a mult. sign where the + sign should be! o_O
 Sci Advisor HW Helper PF Gold P: 3,289 A frustrating limit of a function f(x,y) 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.