Chapter 1.4, Problem 53E

To determine

To Find: If the curve has tangent at the origin.

Answer to Problem 53E

Yes, the graph of the function has a tangent at the origin and the slope is equal to 0.

Explanation of Solution

Given information:

The curve is: $f\left(x\right)=\left\{\begin{array}{l}{x}^2\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}\right.$ and the point is $x=0$ .

Concept used:

If the curve has a tangent at any point say $x=a$, then the two-sided limit of the difference quotient should exist, that is:

  $\underset{h\to 0^{-}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

For any function f , slope at any point $x=a$ can be calculated as: $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Calculation:

First check if the two-sided quotient limit of $f\left(x\right)=\left\{\begin{array}{l}{x}^2\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}\right.$ at $x=0$ exist or not.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{\left({\left(0+h\right)}^2\sin \left(\frac{1}{0+h}\right)+0\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{\left(h^2\sin \left(\frac{1}{h}\right)\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }h\sin \left(\frac{1}{h}\right)\\ =\underset{h\to 0^{-}}{\lim }\frac{\sin \left(\frac{1}{h}\right)}{\left(\frac{1}{h}\right)}\\ =0\\ \underset{h\to 0^{+}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{\left({\left(0+h\right)}^2\sin \left(\frac{1}{0+h}\right)+0\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{\left(h^2\sin \left(\frac{1}{h}\right)\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }h\sin \left(\frac{1}{h}\right)\\ =\underset{h\to 0^{+}}{\lim }\frac{\sin \left(\frac{1}{h}\right)}{\left(\frac{1}{h}\right)}\\ =0\end{array}$

Here it can be seen that $\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ . Hence the tangent at $x=0$ exist.

For any function f , slope at any point $x=0$ can be calculated as: $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=0$ .

Conclusion:

The function has a tangent with slope 0 at the origin.