# How do you use the definition of continuity and the properties of limits to show the function is continuous F(x)= (x^2-8)^8 on the interval (-inf, inf)?

Feb 9, 2017

see below

#### Explanation:

Let a be any number from the interval $\left(- \infty , \infty\right)$ then we need to show that $f \left(a\right) = {\lim}_{x \to a} f \left(x\right)$.

Since $f \left(x\right) = {\left({x}^{2} - 8\right)}^{8}$ then

$f \left(a\right) = {\left({a}^{2} - 8\right)}^{8}$

${\lim}_{x \to a} {\left({x}^{2} - 8\right)}^{8} = {\left({\lim}_{x \to a} {x}^{2} - {\lim}_{x \to a} 8\right)}^{8} = {\left({a}^{2} - 8\right)}^{8}$

Since $f \left(a\right) = {\lim}_{x \to a} f \left(x\right) = {\left({a}^{2} - 8\right)}^{8}$,f is continuous at x=a for every a in $\left(- \infty , \infty\right)$

OR

We can look at $f \left(x\right) = {\left({x}^{2} - 8\right)}^{8}$ and recognize that it is a polynomial and since polynomials are continuous everywhere therefore $f \left(x\right) = {\left({x}^{2} - 8\right)}^{8}$ is continuous everywhere.