
#1
Feb2713, 10:51 PM

P: 517

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? 



#2
Feb2713, 11:16 PM

Sci Advisor
HW Helper
PF Gold
P: 2,903

Maybe try this inequality, valid for ##y \leq \pi##. Plot the two functions to see why it's true.
$$\left\frac{\sin(y)}{y}\right \geq \left1  \frac{1}{\pi}y\right$$ 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. 



#3
Feb2713, 11:48 PM

P: 71

[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




#4
Feb2713, 11:57 PM

Sci Advisor
HW Helper
PF Gold
P: 2,903

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. 


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