Question

Help please! Worth some points!

Answer 1

$f(x)=\begin{cases}5-x^2&\text{for }x0\end{cases}$

$\displaystyle\lim_{x\to0^-}f(x)=\lim_{x\to0}(5-x^2)=5$
$\displaystyle\lim_{x\to0^+}f(x)=\lim_{x\to0}(2x+5)=5$
$\implies\displaystyle\lim_{x\to0}f(x)=5$

- - -

$f(x)=\begin{cases}x-3&\text{for }x4\end{cases}$

$\displaystyle\lim_{x\to4^-}f(x)=\lim_{x\to4}(x-3)=1$
$\displaystyle\lim_{x\to4^+}f(x)=\lim_{x\to4}(x+3)=7$
$\implies\displaystyle\lim_{x\to4}f(x)\text{ does not exist}$