Question

f(x) = \

$$\begin{cases} x + 3 & \\text{for } x < 1 \\\\ -2x + 7 & \\text{for } x > 1 \\end{cases}$$

if $f$ is the function defined above, then $\lim_{x \to 1} f(x)$ is
a) 2
b) 4
c) 5
d) nonexistent

Explanation:

Step1: Find left - hand limit

For \(x < 1\), the function is \(f(x)=x + 3\). The left - hand limit as \(x\to1^{-}\) is \(\lim_{x\to1^{-}}f(x)=\lim_{x\to1^{-}}(x + 3)\). Substitute \(x = 1\) into \(x+3\), we get \(1 + 3=4\).

Step2: Find right - hand limit

For \(x>1\), the function is \(f(x)=-2x + 7\). The right - hand limit as \(x\to1^{+}\) is \(\lim_{x\to1^{+}}f(x)=\lim_{x\to1^{+}}(-2x + 7)\). Substitute \(x = 1\) into \(-2x + 7\), we get \(-2\times1+7=-2 + 7 = 5\).

Step3: Compare left and right limits

Since \(\lim_{x\to1^{-}}f(x)=4\) and \(\lim_{x\to1^{+}}f(x)=5\), and \(4
eq5\), the two - sided limit \(\lim_{x\to1}f(x)\) does not exist.

Answer:

D. nonexistent