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  • Examine the differentiability of f, where f is defined by f(x) = x^2sin 1/x if x not = 0

Examine the differentiability of f, where f is defined by

f(x)=\left\{\begin{array}{cl} x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right. \text { at } x=0

Answers (1)

Given,

f(x)=\left\{\begin{array}{cl} x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right. \text { at } x=0

We need to check whether f(x) is continuous and differentiable at x = 0
A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)

Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).
Mathematically we can represent it as-

\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c}

\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{c-h-c}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{c+h-c}

Finally, we can state that for a function to be differentiable at x = c

\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{-h}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}

Checking for the continuity:

Now according to above theory-

f(x) is continuous at x = 0 if -
\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0) \\ \therefore L H L=h \rightarrow 0 \\ \Rightarrow L H L=\lim _{h \rightarrow 0}\left\{(-h)^{2} \sin \left(\frac{1}{-h}\right)\right\}_{\{u \operatorname{sing}} \text { equation } \left.1\right\}

As sin (-1/h) is going to be some finite value from -1 to 1 as h→0

∴ LHL = 0^2 × (finite value) = 0

∴ LHL = 0 …(2)

Similarly,

\lim _{\mathrm{RHL}}=\lim _{h \rightarrow 0} \mathrm{f}(\mathrm{h}) \\ \Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\left\{(\mathrm{~h})^{2} \sin \left(\frac{1}{\mathrm{~h}}\right)\right\}_{\{\text {using equation } 1\}}

As sin (1/h) is going to be some finite value from -1 to 1 as h→0

∴ RHL = 0^2(finite value) = 0 …(3)

And, f(0) = 0 {using equation 1} …(4)

From equation 2,3 and 4 we observe that:

\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)

∴ f(x) is continuous at x = 0. So we will proceed now to check the differentiability.

Checking for the differentiability:

Now according to above theory-

f(x) is differentiable at x = 0 if -

\\ \lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ \therefore L H D=\lim_{h \rightarrow 0 }\frac{f(-h)-f(0)}{-h} \\ \Rightarrow L H D=\lim_{h \rightarrow 0 }\frac{(-h)^{2} \sin \left(\frac{1}{-h}\right)-0}{-h} \quad\{\text { using equation } 1\} \\ \Rightarrow L H D=\lim _{h \rightarrow 0} \frac{h^{2} \sin \left(\frac{1}{-h}\right)}{-h}=\lim _{h \rightarrow 0}\left\{h \sin \left(\frac{1}{h}\right)\right\}

As sin (1/h) is going to be some finite value from -1 to 1 as h→0

∴ LHD = 0×(some finite value) = 0

∴ LHD = 0 …(5)

Now,

\operatorname{RHD}=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ \Rightarrow R H D=h \rightarrow 0 \frac{(h)^{2} \sin \left(\frac{1}{h}\right)-0}{h} \quad\{\text { using equation } 1\} \\

\Rightarrow \mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0}\left\{\mathrm{~h} \sin \left(\frac{1}{\mathrm{~h}}\right)\right\}

As sin (1/h) is going to be some finite value from -1 to 1 as h→0

∴ RHD = 0×(some finite value) = 0
∴ RHD = 0 …(6)
Clearly from equation 5 and 6, we can conclude that-

(LHD at x=0) = (RHD at x = 0)

∴ f(x) is differentiable at x = 0

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