Question

The slope of the function at x = 0.

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {{x^2}\sin \left( {\frac{1}{x}} \right),}&{x \ne 0}\\ {0,}&{x = 0} \end{array}} \right.\) ?

1. 1
2. 0
3. -1
4. None

Answer 1

Correct Answer - Option 2 : 0

Concept:

The derivative of a function gives us the slope of the line tangent to the function at any point on the graph. 

\(\begin{array}{l} \rm \frac{dy}{dx}=\lim _{x \rightarrow 0} \frac{f(x)-f(a)}{x-a} \\ \end{array}\)

 

Calculation:

Here,  \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {{x^2}\sin \left( {\frac{1}{x}} \right),}&{x \ne 0}\\ {0,}&{x = 0} \end{array}} \right.\)

Slope = m = \(\begin{array}{l} \rm \frac{dy}{dx}=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} \\ \end{array}\)

\(\begin{array}{l} \rm \lim _{x \rightarrow 0} \frac{x^2 \sin (\frac 1 x)-0}{x} \\ \rm \lim _{x \rightarrow 0} x \sin (\frac 1 x) \end{array}\)

As we know, \(\rm -1 \leq \sin (\dfrac{1}{x}) \leq 1\)

Multiplying by x, we get

\(\Rightarrow \rm -x \leq x\sin (\dfrac{1}{x}) \leq x\)

\(\Rightarrow \displaystyle\lim_{\rm x\rightarrow 0} \rm (-x) \leq \displaystyle\lim_{\rm x\rightarrow 0} \rm x \ sin \left(\dfrac{1}{x}\right) \leq \displaystyle\lim_{\rm x\rightarrow 0} \rm x\)

\(\Rightarrow 0 \leq \displaystyle\lim_{\rm x\rightarrow 0} \rm x^2 \ sin \left(\dfrac{1}{x}\right) \leq 0\)

∴ \(\displaystyle\lim_{\rm x\rightarrow 0} \rm x^2 \ sin \left(\dfrac{1}{x}\right)\)= 0

Hence, option (2) is correct.