Question

let $f(x)=\begin{cases}4x + 14&\text{if }x < - 2\sqrt{x + 38}&\text{if }x>-2\\-2&\text{if }x = - 2end{cases}$ which one of the following rules is violated at $x=-2$ a. $f(a)$ is defined b. $lim_{x
ightarrow a}f(x)=f(a)$ c. $lim_{x
ightarrow a}f(x)$ exists d. none of the above

Explanation:

Step1: Check if \(f(a)\) is defined

When \(a = - 2\), \(f(-2)=-2\), so \(f(a)\) is defined.

Step2: Calculate the left - hand limit

For \(x\lt - 2\), \(f(x)=4x + 14\). Then \(\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(4x + 14)=4\times(-2)+14=6\).

Step3: Calculate the right - hand limit

For \(x\gt - 2\), \(f(x)=\sqrt{x + 38}\). Then \(\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}\sqrt{x + 38}=\sqrt{-2 + 38}=\sqrt{36}=6\). So \(\lim_{x
ightarrow - 2}f(x)=6\).

Step4: Compare with \(f(-2)\)

Since \(\lim_{x
ightarrow - 2}f(x)=6\) and \(f(-2)=-2\), we have \(\lim_{x
ightarrow - 2}f(x)
eq f(-2)\).

Answer:

B. \(\lim_{x
ightarrow a}f(x)=f(a)\)