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  • Find which of the functions is continuous or discontinuous at the indicated points: f(x) =2x ^2 -3x -2 cos 2x / x^2

Find which of the functions is continuous or discontinuous at the indicated points:

f(x)=\left\{\begin{aligned} \frac{2 x^{2}-3 x-2}{x-2}, \text { if } & x \neq 2 \\ 5,\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \text { if } x=2 \end{aligned}\right.

at x = 2

Answers (1)

Given,

f(x)=\left\{\begin{aligned} \frac{2 x^{2}-3 x-2}{x-2}, \text { if } & x \neq 2 \\ 5,\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \text { if } x=2 \end{aligned}\right.

We need to check its continuity at x = 2
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 = 2 if -

\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2)

Then,

\begin{aligned} &\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} \frac{2(2-h)^{2}-3(2-h)-2}{(2-h)-2} \quad\{\text { using equation } 1\}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{2\left(4+\mathrm{h}^{2}-4 \mathrm{~h}\right)-6+3 \mathrm{~h}-2}{-\mathrm{h}}\\ &\Rightarrow \mathrm{lim} \frac{8+2 \mathrm{~h}^{2}-8 \mathrm{~h}-6+3 \mathrm{~h}-2}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{2 \mathrm{~h}^{2}-5 \mathrm{~h}}{-\mathrm{h}}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}(5-2 \mathrm{~h})}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0}(5-2 \mathrm{~h})\\ &\therefore \mathrm{LHL}=5-2(0)=5 \ldots(2) \end{aligned}

Similarly, we proceed for RHL-

\lim _{\mathrm{RHL}=\mathrm{h} \rightarrow 0} \mathrm{f}(2+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{2(2+\mathrm{h})^{2}-3(2+\mathrm{h})-2}{(2+\mathrm{h})-2} \text { [using equation } \left.1\right\}

\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{2\left(4+\mathrm{h}^{2}+4 \mathrm{~h}\right)-6-3 \mathrm{~h}-2}{\mathrm{~h}} \\\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{8+2+8 \mathrm{~h}-6-3 \mathrm{~h}-2}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{2 \mathrm{~h}^{2}+5 \mathrm{~h}}{\mathrm{~h}} \\\Rightarrow \lim _{h \rightarrow 0} \frac{h(5+2 h)}{h}=\lim _{h \rightarrow 0}(5+2 h) \\\therefore \mathrm{RHL}=5+2(0)=5 \ldots(3) \\And, f(2)=5\{ using \ equation \ 1\} \ldots(4)

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

\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2) =5

∴ f(x) is continuous at x = 2

Posted by

infoexpert22

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