Third degree polynomial for $ \sin(x^2+y^2)?$ Let R be an open region containing the point $(x_0,y_0).$ Let f, g, and h be functions defined on R, except possibly at$ (x_0,y_0).$ Suppose that for every $(x,y) \in R$ not equal to $(x_0,y_0)$, we have:
$g(x,y) \leq f(x,y) \leq h(x,y)$.
If 
$\lim_{(x,y)\to(x_0,y_0)} g(x,y) = L = \lim_{(x,y)\to(x_0,y_0)} h(x,y)$
then:
$\lim_{(x,y)\to(x_0,y_0)} f(x,y) = L$
Use the squeeze theorem evaluate $\displaystyle \lim_{(x,y)\to(0,0)} \dfrac{\sin (x^2+y^2) }{x^2+y^2}.$
To apply the Squeeze Thoerem we must select functions g and h. Use the Taylor Polynomials of degree 1 and 3 as natural bounds of $\, \sin u $to fill in the blanks.
$? \leq \sin (x^2+y^2) \leq  ?$

Okay, so what is 3rd degree Taylor Polynomial? I know the Taylor formula for sin(x) is:
$$\sum_{k=0}^n{{(-1)}^k \frac{x^{2k +1}}{(2k+1)!}}.$$
So for the 1st degree Taylor polynomial, my value is:
$(-1)^0 \cdot \frac{ (x^2+y^2)^{2 \cdot 0 +1} } {(2 \cdot 0 + 1)!}  $ = $x^2+y^2$
???
Also, do I have to plug in 0 for x and y, since I'm supposed to evaluate at 0?

Treat $x^2+y^2$ as a single variable.  Recall that 
$$z-\frac16z^3\le \sin(z)\le z$$
for $0<z<1$.  Then, let $z=x^2+y^2$ reveals 
$$(x^2+y^2)-\frac16(x^2+y^2)^3\le \sin(x^2+y^2)\le (x^2+y^2)$$
Dividing by $x^2+y^2$ yields
$$1-\frac16(x^2+y^2)^2\le \frac{\sin(x^2+y^2)}{x^2+y^2}\le 1$$
whereupon applying the squeeze theorem, we have 
$$\lim_{(x,y)\to (0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=1$$
