Question

Oscillation There is another way that a function might fail to be differentiable, and that is by oscillation. Let $$f(x)=\left\\{\begin{array}{ll}{x \sin \frac{1}{x},} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right.$$ (a) Show that \(f\) is continuous at \(x=0\) (b) Show that $$\frac{f(0+h)-f(0)}{h}=\sin \frac{1}{h}$$ (c) Explain why $$\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$$ does not exist. (d) Does \(f\) have either a left-hand or right-hand derivative at \(x=0 ?\) (e) Now consider the function $$g(x)=\left\\{\begin{array}{ll}{x^{2} \sin \frac{1}{x},} & {x \neq 0} \\\ {0,} & {x=0}\end{array}\right.$$ Use the definition of the derivative to show that \(g\) is differentiable at \(x=0\) and that \(g^{\prime}(0)=0\)

Short answer

The function \(f\) is continuous at \(x=0\) but not differentiable at \(x=0\) because the derivative oscillates between -1 and 1 as \(h\) approaches 0. The function \(g\) is both continuous and differentiable at \(x=0\), with \(g'(0) = 0\).

Step-by-step solution

Show \(f\) is continuous at \(x=0\)

We know that \(f(x) = x \sin (1/x)\) for \(x \neq 0\) and \(f(x) = 0\) for \(x = 0\). Hence, \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0\). Therefore, \(f\) is continuous at \(x = 0\).

Show \(f'(x)\)

\(f'(x) = \lim_{h \to 0} ((f(0+h)-f(0))/h) = \lim_{h \to 0} \sin (1/h)\).

Explain why \(\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\) does not exist

The limit does not exist because the function \(sin(1/h)\) oscillates between -1 and 1 infinitely as \(h\) approaches 0. There is no specific value that the limit approaches.

Evaluate if \(f\) has either a left-hand or right-hand derivative at \(x=0\)

Since \(\lim_{h \to 0^-} \sin (1/h)\) and \(\lim_{h \to 0^+} \sin (1/h)\) also oscillate between -1 and 1 as \(h\) approaches zero from the left and right respectively, neither left-hand nor right-hand derivative at \(x=0\) exists.

Show that \(g\) is differentiable at \(x = 0\)

We know that \(g(x) = x^2 \sin (1/x)\) for \(x \neq 0\) and \(g(x) = 0\) for \(x = 0\). Its derivative at \(x=0\) is given by \(g'(0) = \lim_{h \to 0} ((g(0+h)-g(0))/h) = \lim_{h \to 0} h \sin (1/h)\). The value of \(h \sin (1/h)\) is always between -|h| and |h|, and hence as \(h\) approaches zero, \(h \sin (1/h)\) approaches zero. Therefore, the limit exists and equals zero, implying \(g(x)\) is differentiable at \(x = 0\), and \(g'(0) = 0\).