How do you use the definition of continuity and the properties of limits to show that the function f(x)=x^2 + sqrt(7-x) is continuous at a=4?

Feb 9, 2017

see below

Explanation:

$f \left(x\right) = {x}^{2} + \sqrt{7 - x} , a = 4$

We need to show that $f \left(4\right) = {\lim}_{x \to 4} {x}^{2} + \sqrt{7 - x}$

That is,

$f \left(4\right) = 16 + \sqrt{7 - 4} = 16 + \sqrt{3}$

${\lim}_{x \to 4} {x}^{2} + \sqrt{7 - x} = {\lim}_{x \to 4} {x}^{2} + {\lim}_{x \to 4} \sqrt{7 - x}$

= lim_(x->4) x^2+sqrt(lim_(x->4)(7-x)

$= {\lim}_{x \to 4} {x}^{2} + \sqrt{{\lim}_{x \to 4} 7 - {\lim}_{x \to 4} x}$

$= 16 + \sqrt{7 - 4}$

$= 16 + \sqrt{3}$

Since $f \left(4\right) = {\lim}_{x \to 4} {x}^{2} + \sqrt{7 - x} = 16 + \sqrt{3}$ therefore

$f \left(x\right) = {x}^{2} + \sqrt{7 - x}$ is continuous at a = 4