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  • Find which of the functions is continuous or discontinuous at the indicated points: f(x) equals 1 minus cos 2x / x^2

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

f(x)=\left\{\begin{array}{cl} \frac{1-\cos 2 x}{x^{2}}, & \text { if } x \neq 0 \\ 5, & \text { if } x=0 \end{array}\right. \text { at } x=0

 

Answers (1)

Given,

f(x)=\left\{\begin{array}{cl} \frac{1-\cos 2 x}{x^{2}}, & \text { if } x \neq 0 \\ 5, & \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 present it as

\lim _{h \rightarrow 0} \mathrm{f}(\mathrm{c}-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(\mathrm{c}+\mathrm{h})=\mathrm{f}(\mathrm{c}) \\ \text{Where h is a very small number very close to 0} (\mathrm{~h} \rightarrow 0) \\ \text{Now according to above theory}- \\ f(x) \text{is continuous at} \ x=0 \ if - \\ \lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)

\begin{aligned} &\text { then, }\\ &\lim _{\mathrm{h} \mathrm{HL}=\mathrm{h} \rightarrow 0} \mathrm{f}(-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2(-\mathrm{h})}{(-\mathrm{h})^{2}} \underbrace{\{\text { using equation } 1\}}\\ &\text { As we know } \cos (-\theta)=\cos \theta\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2 \mathrm{~h}}{\mathrm{~h}^{2}}\\ &\because 1-\cos 2 x=2 \sin ^{2} x\\ &\therefore \mathrm{lim}_{\mathrm{h} \rightarrow 0} \frac{2 \sin ^{2} \mathrm{~h}}{\mathrm{~h}^{2}}=2 \lim _{\mathrm{h} \rightarrow 0}\left(\frac{\sin \mathrm{h}}{\mathrm{h}}\right)^{2} \end{aligned}

As this limit can be evaluated directly by putting value of h because it is taking indeterminate form (0/0)

As we know,

\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\\ &\therefore \mathrm{LHL}=2 \times 1^{2}=2 \ldots(2)\\ &\text { Similarly, we proceed for RHL- }\\ &\lim _{\mathrm{RHL}} \mathrm{h} \rightarrow 0^+{\mathrm{f}}(\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2(\mathrm{~h})}{(\mathrm{h})^{2}}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2 \mathrm{~h}}{\mathrm{~h}^{2}}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{2 \sin ^{2} \mathrm{~h}}{\mathrm{~h}^{2}}=2 \lim _{\mathrm{h} \rightarrow 0}\left(\frac{\sin \mathrm{h}}{\mathrm{h}}\right)^{2} \end{aligned}

Again, using sandwich theorem, we get -

RHL = 2 \times 1^2 = 2...(3)

And,
f (0) = 5 …(4)
From equation 2, 3 and 4 we can conclude that

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

∴ f(x) is discontinuous at x = 0

Posted by

infoexpert22

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