[tex]\begin{gathered} \lim_{x\to-5}f(x) \\ \\ f(x)=\begin{cases}x^2+8x+17,x\leq-5{} \\ -\frac{x}{2}+4,x>-5{}\end{cases} \end{gathered}[/tex]
Evaluate each limit separately:
[tex]\begin{gathered} \lim_{x\to-5}x^2+8x+17 \\ \\ \lim_{x\to-5}(-5)^2+8(-5)+17=25-40+17=2 \\ \\ \end{gathered}[/tex][tex]\begin{gathered} \lim_{x\to-5}-\frac{x}{2}+4 \\ \\ \lim_{x\to-5}-\frac{-5}{2}+4=\frac{5}{2}+4=\frac{5+8}{2}=\frac{13}{2}=6.5 \end{gathered}[/tex]
As the one-sides limits are different, the limit doesn't exist
Answer: Does not exist.
