Question

let ( f(x) = \begin{cases} x^2 + 4, & x < -4 \\sqrt{x + 4}, & x geq -4 end{cases} ) compute the following limits or state that they do not exist.

a. ( limlimits_{x \to -4^-} f(x) )

b. ( limlimits_{x \to -4^+} f(x) )

c. ( limlimits_{x \to -4} f(x) )

a. ( limlimits_{x \to -4^+} f(x) = 0 ) (simplify your answer)

b. the limit does not exist

c. compute the limit of ( limlimits_{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.

a. yes, ( limlimits_{x \to -4} f(x) ) exists and equals \\( \square \\) (simplify your answer.)

b. no, ( limlimits_{x \to -4} f(x) ) does not exist because ( limlimits_{x \to -4^+} f(x) \
eq limlimits_{x \to -4^-} f(x) ).

c. no, ( limlimits_{x \to -4} f(x) ) does not exist because ( f(-4) ) is undefined.

Explanation:

Part a: $\lim_{x \to -4^-} f(x)$

Step1: Identify the function for $x < -4$

For $x \to -4^-$, we use $f(x) = x^2 + 4$.

Step2: Substitute $x = -4$

Substitute $x = -4$ into $x^2 + 4$: $(-4)^2 + 4 = 16 + 4 = 20$.

Step1: Identify the function for $x \geq -4$

For $x \to -4^+$, we use $f(x) = \sqrt{x + 4}$.

Step2: Substitute $x = -4$

Substitute $x = -4$ into $\sqrt{x + 4}$: $\sqrt{-4 + 4} = \sqrt{0} = 0$.

Step1: Compare left and right limits

We found $\lim_{x \to -4^-} f(x) = 20$ and $\lim_{x \to -4^+} f(x) = 0$. Since $20
eq 0$, the two - sided limit does not exist because the left - hand limit and the right - hand limit are not equal.

Answer:

$\lim_{x \to -4^-} f(x) = 20$

Part b: $\lim_{x \to -4^+} f(x)$