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  • f(x) =|x| cos 1/x if x not =0 at x = 0

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

at x = 0

Answers (1)

Given,

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

We need to check its continuity 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)

Now according to above theory-
f(x) is continuous at x = 4 if -

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

then, \\\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{|-\mathrm{h}| \cos \frac{1}{(-\mathrm{h})}\right\}_{\{\mathrm{using} \text { equation } 1\}} \\\because \mathrm{h}>0 \ as \ defined \ above.\\ \\\therefore|-h|=h \\\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}\left\{\mathrm{~h} \cos \left(\frac{1}{\mathrm{~h}}\right)\right\}

As cos (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{h} \rightarrow 0}{\mathrm{f}}(\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{|\mathrm{~h}| \cos \frac{1}{(\mathrm{~h})}\right\}_{\{\text {using equation } 1\}}

\because \mathrm{h}>0 \ as \ defined \ above. \\\therefore|h|=h \\\Rightarrow \lim _{\mathrm{RHL}}=\lim _{h \rightarrow 0}\left\{\mathrm{~h} \cos \left(\frac{1}{\mathrm{~h}}\right)\right\} \\As \ \cos (1 / \mathrm{h}) \ \text{is going to be some finite value from -1 to 1 as} \ \mathrm{h} \rightarrow 0 \\\therefore \mathrm{RHL}=0 \times( finite value )=0 \ldots(3)

And,
f(0) = 0 {using eqn 1} …(4)

 From the above equation 2 , 3 and 4 we can conclude that

\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(4)=0

∴ f(x) is continuous at x = 0

Posted by

infoexpert22

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