Question

Solve this problem
${\large{ \bf{let \: f(x) = \begin{cases}x \sin\dfrac{1}{x} \: \: \: if \: \: x \neq0 \\ 0 \: \: when \: \: x = 0\end{cases}}}}$
Check differentiability and continuity at x = 0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = \begin{cases}x \sin\dfrac{1}{x} \: \: \: if \: \: x \neq0 \\ \\ 0 \: \: when \: \: x = 0\end{cases}$

Case :- 1 To discuss the continuity of f(x) at x = 0

We know,

A function f(x) is said to be continuous at x = a, iff

$\boxed{\sf{ \: \: \rm \: f(a) = \displaystyle\lim_{x \to a ^{ - } }\rm f(x) = \displaystyle\lim_{x \to a^{ + } }\rm f(x) \: \: }}$

Now,

$\rm \: f(0) = 0$

Consider LHL at x = 0

$\rm \: \displaystyle\lim_{x \to 0 ^{ - } }\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to 0 ^{ - } }\rm x \: sin \frac{1}{x}$

To evaluate this limit, we use method of Substitution,

So, Substitute

$\rm \: x = 0 - h, \: \: as \: x \: \to \: 0, \: \: so \: h \: \to \: 0$

So, on substituting this, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm ( - h) \: sin \frac{1}{( - h)}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm h \: sin \frac{1}{ h}$

$\rm \: = \: 0 \times an \: oscillatory \: number \: lies \: between \: - 1 \: and \: 1$

$\rm \: = \: 0$

Now, Consider RHL at x = 0

$\rm \: \displaystyle\lim_{x \to 0 ^{ + } }\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to 0 ^{ + } }\rm x \: sin \frac{1}{x}$

Now, Substitute

$\rm \: x = 0 + h, \: \: as \: x \: \to \: 0, \: \: so \: h \: \to \: 0$

So, using this, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm h \: sin \frac{1}{ h}$

$\rm \: = \: 0 \times an \: oscillatory \: number \: lies \: between \: - 1 \: and \: 1$

$\rm \: = \: 0$

So, from above we concluded that

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

$\rm\implies \:f(x) \: is \: continuous \: at \: x \: = \: 0$

Case :- 2 Now, To check Differentiability at x = 0

We know,

A function f(x) is said to be differentiable at x = a iff

$\boxed{\sf{ \: \: \displaystyle\lim_{x \to a^{ - } }\rm \frac{f(x) - f(a)}{x - a} \: = \: \displaystyle\lim_{x \to a^{ + } }\rm \frac{f(x) - f(a)}{x - a} \: \: }}$

So, from given function, we have

$\rm \: f(0) = 0 \:$

Now, Consider LHD

$\rm \: \displaystyle\lim_{x \to 0^{ - } }\rm \frac{f(x) - f(0)}{x - 0}$

$\rm \: = \: \displaystyle\lim_{x \to 0^{ - } }\rm \frac{x \: sin \dfrac{1}{x} - 0}{x}$

$\rm \: = \: \displaystyle\lim_{x \to 0^{ - } }\rm \frac{x \: sin \dfrac{1}{x}}{x}$

$\rm \: = \: \displaystyle\lim_{x \to 0 ^{ - } }\rm sin \frac{1}{x}$

Now, Substitute

$\rm \: x = 0 - h, \: \: as \: x \: \to \: 0, \: \: so \: h \: \to \: 0$

So, on using this, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm sin \frac{1}{( - h)}$

$\rm \: = \: an \: oscillatory \: number \: lies \: between \: - 1 \: and \: 1$

$\rm\implies \: \displaystyle\lim_{x \to 0^{ - } }\rm \frac{f(x) - f(0)}{x - 0} \: does \: not \: exist$

$\rm\implies \:f(x) \: is \: not \: differentiable \: at \: x \: = \: 0$

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

$\tt\text{To check continuity:} \: \: \[ \underset{x \rightarrow 0} { \lim} \quad \: f(x)=f(0)=0 \]$

Now,

$\[ \begin{array}{l} \tt \underset{x \rightarrow 0}{ \lim} \: \: \: f(x) \\ \\ =\underset{x \rightarrow 0}{ \lim}\left( \tt \: x \sin \left(\frac{1}{x}\right)\right) \\ \\ =0 \end{array} \]$

$\text{Thus, \( \tt f(x) \) is continuous at \( \tt x=0 \).}$

$\text{To check differentiability}$

$\[ \begin{aligned} \tt \: f^{\prime}(0) & \tt=\lim _{x \rightarrow 0}\left(\frac{f(x)-f(0)}{x-0}\right) \\ \\ & \tt=\lim _{x \rightarrow 0}\left(\frac{x \sin \left(\frac{1}{x}\right)-0}{x-0}\right) \\ \\ & \tt=\lim _{x \rightarrow 0}\left(\sin \left(\frac{1}{x}\right)\right) \\ \\ &=\text { Oscillating a finite value between } \\ &-1 \text { to } 1 \end{aligned} \]$

$\text{Thus, \( \tt f(x) \) is not differentiable at \( \tt x=0 \).}$