Question

a. graph ( f(x) = \begin{cases} 1 - x^2, & x
eq 1 \\ 2, & x = 1 end{cases} )
b. find ( limlimits_{x \to 1^-} f(x) ) and ( limlimits_{x \to 1^+} f(x) ).
c. does ( limlimits_{x \to 1} f(x) ) exist? if so, what is it? if not, why not?
b. find ( limlimits_{x \to 1^-} f(x) ). select the correct choice below and, if necessary, fill in the answer box in your choice.
a. ( limlimits_{x \to 1^-} f(x) = ) (simplify your answer.)
b. the limit does not exist.

Explanation:

Step1: Recall left - hand limit definition

For the left - hand limit as \(x
ightarrow1^{-}\), we consider the values of \(x\) that are less than 1 and approach 1. When \(x
eq1\), the function \(f(x)=1 - x^{2}\). So we use the formula \(f(x)=1 - x^{2}\) to find \(\lim_{x
ightarrow1^{-}}f(x)\).

Step2: Substitute \(x = 1\) into \(1 - x^{2}\)

We substitute \(x = 1\) into the function \(y = 1-x^{2}\). Using the formula \(\lim_{x
ightarrow a}g(x)=g(a)\) when \(g(x)\) is a polynomial (and polynomials are continuous everywhere), we have \(\lim_{x
ightarrow1^{-}}(1 - x^{2})=1-(1)^{2}\).
Calculating \(1-(1)^{2}\): \(1 - 1=0\).

Answer:

A. \(\lim\limits_{x
ightarrow1^{-}}f(x)=\boxed{0}\)