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  • Prove that the function f defined by f(x) = x / |x| + 2 x^2 remains discontinuous at x=0, regardless the choice of k.

Prove that the function f defined by

f(x)=\left\{\begin{array}{cl} \frac{x}{|x|+2 x^{2}}, & x \neq 0 \\ k, & x=0 \end{array}\right.

remains discontinuous at x=0, regardless the choice of k.

Answers (1)

Given,

f(x)=\left\{\begin{array}{cl} \frac{x}{|x|+2 x^{2}}, & x \neq 0 \\ k, & x=0 \end{array}\right.

We need to prove that f(x) is discontinuous at x = 0 irrespective of the value of k.

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)....(1)

Where h is a very small number very close to 0 (h→0)
Now, we need to prove that f(x) is discontinuous at x = 0 irrespective of the value of k
If we show that,

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

Then there will not be involvement of k in the equation & we can easily prove it.

So let’s take LHL first -

\begin{aligned} &{L H L}=\lim _{h \rightarrow 0} f(0-h)\\ &\Rightarrow \mathrm{LHL}=\lim _{h \rightarrow 0} \frac{(0-\mathrm{h})}{|0-\mathrm{h}|+2(0-\mathrm{h})^{2}}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{|-\mathrm{h}|+2 \mathrm{~h}^{2}}\\ &\because \mathrm{h}>0 \text { as defined in theory above. }\\ &\therefore|-h|=h \end{aligned}

\\\therefore\lim _{ \mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{\mathrm{h}+2 \mathrm{~h}^{2}}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{\mathrm{h}(1+2 \mathrm{~h})} \\\Rightarrow \mathrm{LHL}=\lim _{h \rightarrow 0} \frac{-1}{(1+2 \mathrm{~h})} \\\therefore \mathrm{LHL}=\frac{-1}{1+2(0)}=-1.....(2)
Now Let's find RHL,
\lim _{\mathrm{RHL}}=\operatorname{h}_{\rightarrow 0} \mathrm{f}(0+\mathrm{h})

\begin{aligned} &\Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0} \frac{(0+\mathrm{h})}{|0+\mathrm{h}|+2(0+\mathrm{h})^{2}}\\ &\Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0} \frac{\mathrm{h}}{|\mathrm{h}|+2 \mathrm{~h}^{2}}\\ &\because \mathrm{h}>0 \text { as defined in theory above. }\\ &\therefore|h|=h\\ &\therefore\mathrm{RHL}=\lim _{ h\rightarrow 0}=\frac{\mathrm{h}}{\mathrm{h}+2 \mathrm{~h}^{2}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}}{\mathrm{h}(1+2 \mathrm{~h})} \end{aligned}

\begin{aligned} &\Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0} \frac{1}{(1+2 \mathrm{~h})}\\ &\therefore \mathrm{RHL}=\frac{1}{1+2(0)}=1 ....(3)\end{aligned}

 From the equation 2 and 3, conclude that
LHL ≠ RHL
Hence,
f(x) is discontinuous at x = 0 irrespective of the value of k.

Posted by

infoexpert22

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