Question

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

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{\bf{sin}}\,x}&{{\bf{if}}\,\,x < \frac{\pi }{{\bf{4}}}}\\{{\bf{cos}}\,x}&{{\bf{if}}\,\,\,x \ge \frac{\pi }{{\bf{4}}}}\end{array}} \right.\)

Short answer

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

Step-by-step solution

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

\(f\left( x \right)\) consist of trigonometric functions \(\sin x\) and \(\cos x\), which are continuous in the interval \(\left( { - \infty ,\frac{\pi }{4}} \right)\) and \(\left( {\frac{\pi }{4},\infty } \right)\).

So, the function is continuous in the interval \(\left( { - \infty ,\frac{\pi }{4}} \right) \cup \left( {\frac{\pi }{4},\infty } \right)\).

Find the limit of the function at \(x = \frac{\pi }{{\bf{4}}}\)

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

\(\begin{aligned}\mathop {\lim }\limits_{x \to {{\frac{\pi }{4}}^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {{\frac{\pi }{4}}^ - }} \sin x\\ &= \sin \frac{\pi }{4}\\ &= \frac{1}{{\sqrt 2 }}\end{aligned}\)

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

\(\begin{aligned}\mathop {\lim }\limits_{x \to {{\frac{\pi }{4}}^ + }} f\left( x \right)& = \mathop {\lim }\limits_{x \to {{\frac{\pi }{4}}^ + }} \cos x\\ &= \cos \frac{\pi }{4}\\ &= \frac{1}{{\sqrt 2 }}\end{aligned}\)

The left-hand limit and right-hand limit are equal at \(x = \frac{\pi }{4}\). So, the function is continuous at in the interval \(\left( { - \infty ,\infty } \right)\).