let $f(x)=\begin{cases}4x + 14&\text{if }x < - 2\\sqrt{x + 38}&\text{if }x>-2\\-2&\text{if }x =…

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. $lim_{x\rightarrow a}f(x)=f(a)$ c. $lim_{x\rightarrow a}f(x)$ exists d. none of the above

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. $lim_{x\rightarrow a}f(x)=f(a)$ c. $lim_{x\rightarrow a}f(x)$ exists d. none of the above

Answer

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\rightarrow - 2^{-}}f(x)=\lim_{x\rightarrow - 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\rightarrow - 2^{+}}f(x)=\lim_{x\rightarrow - 2^{+}}\sqrt{x + 38}=\sqrt{-2 + 38}=\sqrt{36}=6). So (\lim_{x\rightarrow - 2}f(x)=6).

Step4: Compare with (f(-2))

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

Answer:

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