4.2 Limits and continuity  (Page 4/14)

There are three conditions to be satisfied, per the definition of continuity. In this example, $a=5$ and $b=\mathrm{-3}.$

1. $f\left(a,b\right)$ exists. This is true because the domain of the function $f$ consists of those ordered pairs for which the denominator is nonzero (i.e., $x+y+1\ne 0).$ Point $\left(5,\mathrm{-3}\right)$ satisfies this condition. Furthermore,
   
   $f\left(a,b\right)=f\left(5,\mathrm{-3}\right)=\frac{3\left(5\right)+2\left(\mathrm{-3}\right)}{5+\left(\mathrm{-3}\right)+1}=\frac{15-6}{2+1}=3.$
2. $\underset{\left(x,y\right)\to \left(a,b\right)}{\text{lim}}f\left(x,y\right)$ exists. This is also true:
   
   $\begin{array}{cc}\hfill \underset{\left(x,y\right)\to \left(a,b\right)}{\text{lim}}f\left(x,y\right)& =\underset{\left(x,y\right)\to \left(5,\mathrm{-3}\right)}{\text{lim}}\frac{3x+2y}{x+y+1}\hfill \\ & =\frac{\underset{\left(x,y\right)\to \left(5,\mathrm{-3}\right)}{\text{lim}}\left(3x+2y\right)}{\underset{\left(x,y\right)\to \left(5,\mathrm{-3}\right)}{\text{lim}}\left(x+y+1\right)}\hfill \\ & =\frac{15-6}{5-3+1}\hfill \\ & =3.\hfill \end{array}$
3. $\underset{\left(x,y\right)\to \left(a,b\right)}{\text{lim}}f\left(x,y\right)=f\left(a,b\right).$ This is true because we have just shown that both sides of this equation equal three.

The polynomials $g\left(x\right)=4x^3$ and $h\left(y\right)=y^2$ are continuous at every real number, and therefore by the product of continuous functions theorem, $f\left(x,y\right)=4x^3{y}^2$ is continuous at every point $\left(x,y\right)$ in the $xy\text{-plane.}$ Since $f\left(x,y\right)=4x^3{y}^2$ is continuous at every point $\left(x,y\right)$ in the $xy\text{-plane}$ and $g\left(x\right)=\text{cos}x$ is continuous at every real number $x,$ the continuity of the composition of functions tells us that $g\left(x,y\right)=\text{cos}\left(4x^3{y}^2\right)$ is continuous at every point $\left(x,y\right)$ in the $xy\text{-plane.}$