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  • Find which of the functions is continuous or discontinuous at the indicated points:Check continuity at x = a f(x) = |x-a| sin 1/|x-a|

Find which of the functions is continuous or discontinuous at the indicated points:
Check continuity at x =a  f(x)=\left\{\begin{array}{cl} |x-a| \sin \frac{1}{x-a}, \text { if } & x \neq a \\ 0, & \text { if } x=a \end{array}\right.

Answers (1)

Given,

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

We need to check its continuity at x = a
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)
Now according to above theory-

f(x) is continuous at x = a if -

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

Clearly,

\begin{aligned} &\left.\lim _{\mathrm{h} \mathrm{HL}} \mathrm{f}(\mathrm{a}-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{|\mathrm{a}-\mathrm{h}-\mathrm{a}| \sin \frac{1}{(\mathrm{a}-\mathrm{h}-\mathrm{a})}\right\}_{\{\text {using }} \operatorname{egn} 1\right\}\\ &\Rightarrow \mathrm{LHL}=\lim _{h \rightarrow 0}\left\{|-\mathrm{h}| \sin \frac{1}{(-\mathrm{h})}\right\}\\ &\because \mathrm{h}>0 \text { as defined above. }\\ &\therefore|-h|=h\\ &\Rightarrow \mathrm{LHL}=\lim _{h \rightarrow 0}\left\{\mathrm{~h} \sin \left(-\frac{1}{\mathrm{~h}}\right)\right\} \end{aligned}

As sin (-1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHL = 0 × (finite value) = 0 …(2)
Similarly we proceed for RHL-

\lim _{\mathrm{RHL}}=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(\mathrm{a}+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{|\mathrm{a}+\mathrm{h}-\mathrm{a}| \sin \frac{1}{(\mathrm{a}+\mathrm{h}-\mathrm{a})}\right\}_{\{\text {using eqn } 1\}}.

 h > 0 as defined above.
∴ |h| = h

\Rightarrow \mathrm{RHL}=\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

∴ RHL = 0 × (finite value) = 0 …(3)
And,

f(a) = 0 {using eqn 1} …(4)

 From equation 2, 3 and 4 we can conclude that

\begin{aligned} &\lim _{h \rightarrow 0} f(a-h)=\lim _{h \rightarrow 0} f(a+h)=f(a)=0\\ &\therefore f(x) \text { is continuous at } x=a \end{aligned}

Posted by

infoexpert22

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