graph the following function and then find the specified limit. when necessary, state that the limit does…

graph the following function and then find the specified limit. when necessary, state that the limit does not exist.\ng(x)=\begin{cases}x^{2},& \text{for }x < 1\\-x + 2,& \text{for }x>1end{cases}\text{ find }lim_{x\rightarrow1}g(x)\nchoose the correct graph below

graph the following function and then find the specified limit. when necessary, state that the limit does not exist.\ng(x)=\begin{cases}x^{2},& \text{for }x < 1\\-x + 2,& \text{for }x>1end{cases}\text{ find }lim_{x\rightarrow1}g(x)\nchoose the correct graph below

Answer

Explanation:

Step1: Calculate left - hand limit

For (x\lt1), (G(x)=x^{2}). So, (\lim_{x\rightarrow1^{-}}G(x)=\lim_{x\rightarrow1^{-}}x^{2}). Substitute (x = 1) into (x^{2}), we get (\lim_{x\rightarrow1^{-}}x^{2}=1^{2}=1).

Step2: Calculate right - hand limit

For (x\gt1), (G(x)=-x + 2). So, (\lim_{x\rightarrow1^{+}}G(x)=\lim_{x\rightarrow1^{+}}(-x + 2)). Substitute (x = 1) into (-x + 2), we get (\lim_{x\rightarrow1^{+}}(-x + 2)=-1 + 2=1).

Step3: Determine the limit

Since (\lim_{x\rightarrow1^{-}}G(x)=\lim_{x\rightarrow1^{+}}G(x)=1), then (\lim_{x\rightarrow1}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\rightarrow1}G(x)=1)