Question

41-42 Show that f is continuous on \(\left( { - \infty ,\infty } \right)\).

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{\bf{1}} - {x^{\bf{2}}}}&{{\bf{if}}\,\,x \le {\bf{1}}}\\{{\bf{ln}}\,x}&{{\bf{if}}\,\,\,x > {\bf{1}}}\end{array}} \right.\)

Short answer

The function is continuous on \(\left( { - \infty ,\infty } \right)\).

Step-by-step solution

Check the function \(f\left( x \right)\) 

For the interval, \(\left( { - \infty ,1} \right)\) the function \(1 - {x^2}\) iscontinuous. Similarly, in the interval, \(\left( {1,\infty } \right)\) the function \(\ln x\) is continuous.

Find the limit of the function at \(x = {\bf{1}}\)

Find the left-hand limitfor \(f\left( x \right)\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {1 - {x^2}} \right)\\ = 1 - 1\\ = 0\end{array}\)

Find theright-hand limit for \(f\left( x \right)\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\ln x} \right)\\ = \ln 1\\ = 0\end{array}\)

As the left-hand limit and right-hand limit are equal at \(x = 1\). So, the function is continuous in the interval \(\left( { - \infty ,\infty } \right)\).