Question

1. based on the graph of y = f(x) below, find

lim_{x\to - 2^{+}}f(x)
lim_{x\to - 2}f(x)
lim_{x\to 2^{-}}f(x)
lim_{x\to 2^{+}}f(x)

Explanation:

Step1: Analyze $\lim_{x\to - 2^{+}}f(x)$

As \(x\) approaches \(-2\) from the right (\(x\to - 2^{+}\)), we look at the part of the graph where \(x>-2\) near \(x =-2\). The value of the function approaches \(2\).

Step2: Analyze $\lim_{x\to - 2}f(x)$

The two - sided limit \(\lim_{x\to - 2}f(x)\) exists if \(\lim_{x\to - 2^{-}}f(x)=\lim_{x\to - 2^{+}}f(x)\). From the graph, when \(x\to - 2^{-}\) (approaching \(-2\) from the left), we consider the part of the graph where \(x < - 2\) near \(x=-2\). The line for \(x < - 2\) is \(y=x + 2\). Substituting \(x=-2\) into \(y=x + 2\), we get \(y=0\). Since \(\lim_{x\to - 2^{-}}f(x)=0\) and \(\lim_{x\to - 2^{+}}f(x)=2\), \(\lim_{x\to - 2}f(x)\) does not exist.

Step3: Analyze $\lim_{x\to 2^{-}}f(x)$

As \(x\) approaches \(2\) from the left (\(x\to 2^{-}\)), we look at the part of the graph where \(x<2\) near \(x = 2\). The line for \(x<2\) (near \(x = 2\)) is \(y=-x\). Substituting \(x = 2\) into \(y=-x\), we get \(y=-2\). So \(\lim_{x\to 2^{-}}f(x)=-2\)

Step4: Analyze $\lim_{x\to 2^{+}}f(x)$

As \(x\) approaches \(2\) from the right (\(x\to 2^{+}\)), we look at the part of the graph where \(x>2\) near \(x = 2\). The line for \(x>2\) (near \(x = 2\)): Let the equation of the line be \(y=mx + b\). Using two points \((2,2)\) (open - circle, but for the limit we consider the trend) and \((0,0)\) (not really, but if we assume a linear function for \(x>2\) with slope \(m = 1\) (since for \(x>2\), if we take a general point \((x,y)\) and assume the line passes through \((2,2)\) (open - circle, but for limit calculation) and say \((3,3)\) (by the trend of the line), the equation is \(y=x\). Substituting \(x = 2\) (for the limit as \(x\to 2^{+}\)), we get \(y = 2\). So \(\lim_{x\to 2^{+}}f(x)=2\)

Answer:

\(\lim_{x\to - 2^{+}}f(x)=2\); \(\lim_{x\to - 2}f(x)\) does not exist; \(\lim_{x\to 2^{-}}f(x)=-2\); \(\lim_{x\to 2^{+}}f(x)=2\)