Question

Sketch the graph of the function \(f\)and use it to determine

the values of \(a\) for which \(\mathop {lim}\limits_{x \to a} f\left( x \right)\) exists.

12. \(f\left( x \right) = \left\{ \begin{aligned}{l}\sqrt({\bf{3}}){x}\;\;\;\;\;\;if\;\;\;\;\;\;x \le - {\bf{1}}\\x\;\;\;\;\;\;\;\;\;if\,\; - {\bf{1}} < x \le 2\\{\left( {x - {\bf{1}}} \right)^{\bf{2}}}\;if\,\;\;\;\;\;x > 2\end{aligned} \right.\)

Short answer

The graph of the function is shown below:

The \(\mathop {lim}\limits_{x \to a} f\left( x \right)\) exists for all \(a \in \left( { - \infty ,2} \right) \cup \left( {2,\infty } \right)\).

Step-by-step solution

Plot the graph of a given piecewise function

For \(x \le - 1\), draw the graph of the function \(\sqrt(3){x}\).For \( - 1 < x \le 2\) draw the graph of function \(x\).And, for \(x > 2\) draw the graph of function \({\left( {x - 1} \right)^2}\).

Therefore, the graph of the function is shown as:

Determine the limits of the function from the graph

It is observed from the graph that the value of the function \(f\left( x \right)\) approaches to \(2\) as \(x\) approaches to \(2\) from the left and value of function \(f\left( x \right)\) approaches to \(1\) as \(x\) approaches to \(2\) from the right. We have \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 2\) and \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 1\).

Since both the left-hand limit and right-hand limit are not equal. This implies that limit of the function at \(x = 0\) does not exist. Therefore, \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = {\rm{does}}\;{\rm{not}}\,{\rm{exist}}\).

This implies that \(\mathop {lim}\limits_{x \to a} f\left( x \right)\) exists for all values of \(a\) except \(2\), that is, \(\mathop {lim}\limits_{x \to a} f\left( x \right)\) exists for all \(a \in \left( { - \infty ,2} \right) \cup \left( {2,\infty } \right)\).