Question

graph the following function and then find the specified limit. when necessary, state that the limit does not exist.
g(x)=\begin{cases}x^{2},& \text{for }x < 1\\-x + 2,& \text{for }x>1end{cases}\text{ find }lim_{x
ightarrow1}g(x)
choose the correct graph below

Explanation:

Step1: Calculate left - hand limit

For \(x\lt1\), \(G(x)=x^{2}\). So, \(\lim_{x
ightarrow1^{-}}G(x)=\lim_{x
ightarrow1^{-}}x^{2}\). Substitute \(x = 1\) into \(x^{2}\), we get \(\lim_{x
ightarrow1^{-}}x^{2}=1^{2}=1\).

Step2: Calculate right - hand limit

For \(x\gt1\), \(G(x)=-x + 2\). So, \(\lim_{x
ightarrow1^{+}}G(x)=\lim_{x
ightarrow1^{+}}(-x + 2)\). Substitute \(x = 1\) into \(-x + 2\), we get \(\lim_{x
ightarrow1^{+}}(-x + 2)=-1 + 2=1\).

Step3: Determine the limit

Since \(\lim_{x
ightarrow1^{-}}G(x)=\lim_{x
ightarrow1^{+}}G(x)=1\), then \(\lim_{x
ightarrow1}G(x)=1\).

To graph the function:

• For \(y = x^{2}\) when \(x\lt1\), it is a part of a parabola opening upwards with the vertex at \((0,0)\) and we consider the values of \(x\) less than \(1\).
• For \(y=-x + 2\) when \(x\gt1\), it is a straight - line with slope \(-1\) and \(y\) - intercept \(2\), and we consider the values of \(x\) greater than \(1\). There is a hole at \(x = 1\) for both parts of the piece - wise function.

Answer:

The limit \(\lim_{x
ightarrow1}G(x)=1\)