Question

let f(x)=\

$$\begin{cases} x^2 + 4, & x < -4 \\\\ \\sqrt{x + 4}, & x \\ge -4 \\end{cases}$$

compute the following limits or state that they do not exist\
a. \\( \lim_{x \to -4^-} f(x) \\)\
b. \\( \lim_{x \to -4^+} f(x) \\)\
c. \\( \lim_{x \to -4} f(x) \\)\
\
a. compute the limit of \\( \lim_{x \to -4^-} f(x) \\) or state that it does not exist. select the correct choice below and, if necessary, fill in the answer box to complete your choice\
\
\\( \bigcirc \\) a. \\( \lim_{x \to -4^-} f(x) = \square \\) (simplify your answer)\
\\( \bigcirc \\) b. the limit does not exist

Explanation:

Step1: Identify the function for left - hand limit

For the left - hand limit as \(x
ightarrow - 4^{-}\), we use the part of the piece - wise function where \(x\lt - 4\), which is \(f(x)=x^{2}+4\).

Step2: Substitute \(x = - 4\) into the function

We calculate the limit \(\lim_{x
ightarrow - 4^{-}}f(x)=\lim_{x
ightarrow - 4^{-}}(x^{2}+4)\).
Substitute \(x=-4\) into \(x^{2}+4\): \((-4)^{2}+4=16 + 4=20\).

Answer:

A. \(\lim_{x
ightarrow - 4^{-}}f(x)=\boxed{20}\)