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.

11.\(f\left( x \right) = \left\{ \begin{aligned}{l}{e^x}\;\;\;\;\;\;\;if\;\;x \le {\bf{0}}\\x - 1\;\;\;\;if\,\;\;{\bf{0}} < x < {\bf{1}}\\lnx\;\;\;\;\;\,if\,\;\;x \ge {\bf{1}}\end{aligned} \right.\)

Short answer

The graph of the function f is shown below:

The \(\mathop {lim}\limits_{x \to a} f\left( x \right)\) exists for all \(a \in \left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\).

Step-by-step solution

Plot the graph of given piecewise function.

For \(x \le 0\), draw the graph of function \({e^x}\).For \(0 < x < 1\) draw the graph of function \(x - 1\).And, for \(x \ge 1\) draw the graph of function \(\ln x\).

Therefore, the graph of function is shown as:

Obtain 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 \(1\) as \(x\) approaches to \(0\) from the left and value of function \(f\left( x \right)\) approaches to \( - 1\) as \(x\) approaches to \(0\) from the right.

So,\(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 1\) and \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - 1\).

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, \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = {\rm{does}}\;{\rm{not}}\,{\rm{exist}}\).

Therefore,\(\mathop {lim}\limits_{x \to a} f\left( x \right)\) exists for all values of \(a\) except \(0\), that is, \(\mathop {lim}\limits_{x \to a} f\left( x \right)\) exists for all \(a \in \left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\).