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The function $f\left( x \right)=x{{\tan }^{-1}}\dfrac{1}{x}$ for $x\ne 0$ , $f\left( 0 \right)=0$ is
A. Differentiable at x = 0
B. neither continuous at x = 0 nor differentiable at x = 0
C. Not continuous at x = 0
D. continuous at x = 0 but not differentiable at x = 0

Last updated date: 23rd May 2024
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Hint: We need to find whether the function $f\left( x \right)=x{{\tan }^{-1}}\dfrac{1}{x}$ is differentiable and continuous at x = 0. We solve the given question by finding out the range of the given function and apply limits to get the desired result.

Complete step-by-step solution:
We are given a function $f\left( x \right)=x{{\tan }^{-1}}\dfrac{1}{x}$ and are asked to find if the function is differentiable and continuous at x = 0. We will be solving the given question using the concept of limits and
We will have to find the range of the given function $f\left( x \right)=x{{\tan }^{-1}}\dfrac{1}{x}$ .
From the formulae of trigonometry, we know that the range of the function ${{\tan }^{-1}}\dfrac{1}{x}$ is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
Writing the same in the form of inequality, we get,
$\Rightarrow -\dfrac{\pi }{2} < {{\tan }^{-1}}\dfrac{1}{x} < \dfrac{\pi }{2}$
Multiplying the inequality with x, we get,
$\Rightarrow -\dfrac{\pi }{2}x < x{{\tan }^{-1}}\dfrac{1}{x} < x\dfrac{\pi }{2}$
A function $f\left( x \right)$ is continuous if, for every value of c in the domain, the function $f\left( c \right)$ is defined and
$\Rightarrow \displaystyle \lim_{x \to c}f\left( x \right)=f\left( c \right)$
We need to find the limit of the function $f\left( x \right)=x{{\tan }^{-1}}\dfrac{1}{x}$ .
Following the same, we get,
$\Rightarrow \displaystyle \lim_{x \to 0}f\left( x \right)=\displaystyle \lim_{x \to 0}x{{\tan }^{-1}}\dfrac{1}{x}$
Applying the limits to the above equation, we get,
$\Rightarrow \displaystyle \lim_{x \to 0}f\left( x \right)=\displaystyle \lim_{x \to 0}x{{\tan }^{-1}}\dfrac{1}{x}=0$
From the above,
$\therefore \displaystyle \lim_{x \to 0}x{{\tan }^{-1}}\dfrac{1}{x}=0$
And from the question we know that $f\left( 0 \right)=0$
$\therefore f\left( x \right)$ is continuous at x = 0.
We can test if the function is differentiable at any point c using the limit given as follows,
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{f\left( c+h \right)-f\left( c \right)}{h}$
If the limit exists, the function is differentiable.
As per the given question, we get,
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{f\left( x \right)-f\left( 0 \right)}{x-0}$
Substituting the values in the above expression, we get,
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x{{\tan }^{-1}}\dfrac{1}{x}-0}{x}$
Simplifying the above equation, we get,
$\Rightarrow \displaystyle \lim_{x \to 0}{{\tan }^{-1}}\dfrac{1}{x}$
We know that the value of the limit ${{\tan }^{-1}}\dfrac{1}{x}$ does not exist at $x=0$ . So, the function is not differentiable at x = 0.
$\therefore$ Option D holds the correct answer for the given question.

Note: We must know the basic formulae of derivatives and limits to solve the given question easily.
The continuity of the function given in the question can also be determined by plotting the graph of
the function. If the graph of the function has no holes or breaks in it the function is continuous.