I attached the question. Could someone help me get started? I can never seem to wrap my head around how to solve piecewise functions for some reason.
Thanks
$f(x) = \begin{cases}\dfrac{x-9}{\sqrt{x}-3},&x \neq 9 \\ a, &x=9 \end{cases}$
in order for $f(x)$ to be continuous the value at $x=9$ must equal the $\displaystyle{\lim_{x\to 9}}~f(x)$
We can quickly apply L'Hopital's rule (sorry Archie) to see
$\underset{x \to 9}{\lim}~f(x) =\left . \dfrac{1}{\dfrac{1}{2\sqrt{x}}} \right |_{x=9}=\dfrac{1}{\dfrac{1}{6}}=6$
so setting $a=6$ makes $f(x)$ continuous at $x=9$
A function is continuous at $x=c$ if $\displaystyle \lim_{x \to c} f(x) = f(c)$
In this case, $c=9$, and you want $f(9)=a$, where
$\displaystyle a = \lim_{x \to 9} \dfrac{x-9}{\sqrt{x}-3}$
Note that $\dfrac{x-9}{\sqrt{x}-3}=\dfrac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}-3}$
Notice that you should add that "for all values of x except x= 9, $\displaystyle \frac{(\sqrt{x}- 3)(\sqrt{x}+ 3)}{\sqrt{x}- 3}= \sqrt{3}+ 3$. Fortunately, the "limit of f(x) as x goes to 9" does not depend upon the value of f at x= 9.