Question

Find $\underset{x\to 5}{lim}f\left(x\right)$ where $f\left(x\right)=\{\begin{array}{c}x+4\\ 5\\ {(x-2)}^2\end{array}\begin{array}{c}x<5\\ x=5\\ x>5\end{array}$

Answer 1

${lim}_{x\to 5^{-}}f\left(x\right)={lim}_{x\to 0}f(5-h)={lim}_{x\to 0}5-h+4=9$

${lim}_{x\to 5^{+}}f\left(x\right)={lim}_{x\to 0}f(5+h)={lim}_{x\to 0}(5+h-2{)}^2=9$

$\to {lim}_{x\to 5}f\left(x\right)=9$