Question

Let \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{{\left| x \right|}^x}}&{if\;x \ne {\bf{0}}}\\{\bf{1}}&{if\;x = {\bf{0}}}\end{array}} \right.\)

(a) Show that \(f\) is continuous at \({\bf{0}}\).

(b) Investigate graphically whether \(f\) is differentiable at \({\bf{0}}\) by zooming in several times toward the point \(\left( {{\bf{0}},{\bf{1}}} \right)\) on the graph of \(f\).

(c) Show that \(f\) is not differentiable at \({\bf{0}}\). How can you reconcile this fact with the appearance of the graphs in part (b)?

Short answer

(a) It is proved that \(f\) is continuous at \(0\).

(b) It is proved that \(f\) is differentiable at \(0\).

(c) It is proved that \(f\) is not differentiable at \(0\).

Step-by-step solution

(a) Step 1: Show that \(f\) is continuous at \({\bf{0}}\).

To show \(f\) is continuous we need the limit of the function as:

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

For \(x \ne 0\) then we have,

\(\begin{array}{c}\ln f\left( x \right) = \ln {\left| x \right|^x}\\ = x\ln \left| x \right|\end{array}\)

Therefore,

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

Thus,

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

Hence \(f\) is continuous at \(0\).

(b) Step 2: Investigate graphically whether \(f\) is differentiable at \({\bf{0}}\) by zooming in several times toward the point \(\left( {{\bf{0}},{\bf{1}}} \right)\) on the graph of \(f\).

Construct the graph for the function \(f\).

On observing the graph it is proved that \(f\) is differentiable at \(0\).

Hence Proved.

Step 3: Show that \(f\) is not differentiable at \({\bf{0}}\). How can you reconcile this fact with the appearance of the graphs in part (b)?

Use logarithmic differentiation to find \(f'\)

\(\begin{array}{c}\ln f\left( x \right) = x\ln \left| x \right|\\\frac{{f'\left( x \right)}}{{f\left( x \right)}} = x\left( {\frac{1}{x}} \right) + \ln \left| x \right|\\f'\left( x \right) = f\left( x \right)\left( {1 + \ln \left| x \right|} \right)\\ = {\left| x \right|^x}\left( {1 + \ln \left| x \right|} \right),\;x \ne 0\end{array}\)

Now \(f'\left( x \right) \to \infty \) so the curve has a vertical tangent at \(\left( {0,1} \right)\) and it is not differentiable there.

The fact cannot be seen in the graphs in part (b) because ln \(\ln \left| x \right| \to - \infty \) very slowly as \(x \to \infty \).

Hence Proved.