Chapter 2.4, Problem 51E

To determine

To check: Whether the graph of the function $f\left(x\right)=\{\begin{array}{l}{x}^2\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}$ has a tangent at origin or not.

Answer to Problem 51E

Yes, the graph of the function $f\left(x\right)=\{\begin{array}{l}{x}^2\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}$ has a tangent at origin.

Explanation of Solution

Given information:

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

Calculation:

Check the slope of the function at $x=0$ to find the tangent at origin.

The slope of the tangent at point can be calculated as:

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

Substitute $h$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=x^2\sin \left(\frac{1}{x}\right)\\ f\left(h\right)=h^2\sin \left(\frac{1}{h}\right)\end{array}$

The value of the function $f\left(x\right)$ is equal to $0$ at $x=0$ . So, the value of $f\left(0\right)$ is equal to $0$.

Substitute $h^2\sin \left(\frac{1}{h}\right)$ for $f\left(h\right)$ and $0$ for $f\left(0\right)$ in the difference quotient.

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

As the trigonometric function $\sin x$ lies between $-1$ to $1$ . So, the function $h\sin \left(\frac{1}{h}\right)$ lies between $-h$ and $h$.

The value of $\underset{h\to 0}{\lim }\left(-h\right)=\underset{h\to 0}{\lim }\left(h\right)$ both are equal to $0$ . So, by the help of sandwich theorem the function $h\sin \left(\frac{1}{h}\right)$ tends to $0$ as $h$ tends to $0$.

The slope of the curve $f\left(x\right)=\{\begin{array}{l}{x}^2\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}$ at $x=0$ is equal to $0$.

Therefore, the graph of the curve $f\left(x\right)=\{\begin{array}{l}{x}^2\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}$ has a tangent at origin.