Question

$g(x) = \

$$\begin{cases} 2x + 3, & x > 4 \\\\ -2x + 3, & x \\leq 4 \\end{cases}$$

$ what is the limit as x approaches 4? options: does not exist, 11, 3, -5

Explanation:

Step1: Find the right - hand limit (x approaches 4 from the right)

When \(x>4\), the function is \(g(x)=2x + 3\). To find the right - hand limit as \(x\to4^{+}\), we substitute \(x = 4\) into the function \(2x+3\).
\(\lim_{x\to4^{+}}g(x)=\lim_{x\to4^{+}}(2x + 3)=2\times4+3=8 + 3=11\)

Step2: Find the left - hand limit (x approaches 4 from the left)

When \(x\leq4\), the function is \(g(x)=- 2x+3\). To find the left - hand limit as \(x\to4^{-}\), we substitute \(x = 4\) into the function \(-2x + 3\).
\(\lim_{x\to4^{-}}g(x)=\lim_{x\to4^{-}}(-2x + 3)=-2\times4+3=-8 + 3=-5\)

Step3: Compare the left - hand and right - hand limits

For the limit as \(x\to4\) to exist, the left - hand limit and the right - hand limit must be equal. Here, \(\lim_{x\to4^{-}}g(x)=-5\) and \(\lim_{x\to4^{+}}g(x)=11\), and \(-5
eq11\). So the limit as \(x\) approaches 4 does not exist.

Answer:

Does Not Exist