Question

If

f (x) = \{\begin{matrix} (\frac{e^{|x|} + | x | - 1}{[x] + \{2 x\}}) & x \neq 0 \\ \frac{1}{2} & x = 0 \end{matrix}

, comment on the continuity at

x = 0

and differentiability at

x = 0

, where

[.]

is the greatest integer function and

\{.\}

is the fractional part function

Accepted Answer

Let us first find out whether the function $f (x)$ is continuous at $x = 0$ or not. Since a function which is discontinuous can never be differentiable at a particular point.

We know that if a function is continuous at $x = 0$ , then

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0)$

Evaluating required limits,

$\lim_{x \to 0^{+}} |x| = x$ ,

$\lim_{x \to 0^{+}} [x] = 0$ ,

$\lim_{x \to 0^{+}} \{2 x\} = 2 x,$ $($ where $[.]$ and $\{.\}$ denote the integral and fractional part of $f (x)$ $)$

Using the above limits,

R.H.L of $f (x)$ at $x = 0$

$\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} \frac{e^{x} + x - 1}{0 + 2 x}$

$= \lim_{x \to 0^{+}} \frac{e^{x} - 1}{2 x} + \frac{x}{2 x}$ $. . . (i)$

Again we know that series expansion of $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + . . .$

Putting this value in $(i)$ , we get

$\lim_{x \to 0^{+}} \frac{(1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . . .) - 1}{2 x} + \frac{x}{2 x}$

$= \lim_{x \to 0^{+}} \frac{x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . . .}{2 x} + \frac{1}{2}$

$= \lim_{x \to 0^{+}} \frac{x (1 + \frac{x}{2!} + \frac{x^{2}}{3!} + . . . . . . .)}{2 x} + \frac{1}{2}$

$= \lim_{x \to 0^{+}} \frac{1}{2} (1 + \frac{x}{2!} + \frac{x^{2}}{3!} + . . . . . . .) + \frac{1}{2}$

$= \frac{1}{2} (1) + \frac{1}{2}$ $= 1$

So R.H.L $= 1$

And it's given in the question that $f (0) = \frac{1}{2}$

$∴$ R.H.L $\neq f (0)$

So $f (x)$ is discontinuous hence, non-differentiable at $x = 0$ .

Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 1: Exercise-1

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 1: Exercise-1

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