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  • Find the value of k so that the function f is continuous at the indicated point: f(x) = 1-coskx/xsinx

Find the value of k so that the function f is continuous at the indicated point:

\begin{array}{c} f(x)=\frac{1-\cos \mathrm{kx}}{\mathrm{x} \sin \mathrm{x}}, \quad \text { if } \mathrm{x} \neq 0 \\ \frac{1}{2}, \quad \text { if } \mathrm{x}=0 \end{array}  at x= 0

Answers (1)

Given,

\begin{array}{c} f(x)=\frac{1-\cos \mathrm{kx}}{\mathrm{x} \sin \mathrm{x}}, \quad \text { if } \mathrm{x} \neq 0 \\ \frac{1}{2}, \quad \text { if } \mathrm{x}=0 \end{array}

We need to find the value of k such that f(x) is continuous 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, let’s assume that f(x) is continuous at x = 0.

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

to find k we have to pick out a combination so that we get k in our equation.
In this question we take LHL = f(0)

\begin{array}{l} \lim _{h \rightarrow 0} f(-h)=f(0) \\ \quad \lim _{h \rightarrow 0}\left\{\frac{1-\cos k(-h)}{(-h) \sin (-h)}\right\}=\frac{1}{2}\{u \text { ing equation } 1\} \end{array}

\begin{array}{l} \because \cos (-x)=\cos x \text { and } \sin (-x)=-\sin x \\ \quad \lim _{h \rightarrow 0}\left\{\frac{1-\cos k(h)}{(h) \sin (h)}\right\}=\frac{1}{2} \\ \text { Also, } 1-\cos x=2 \sin ^{2}(x / 2) \\ \quad \lim _{h \rightarrow 0}\left\{\frac{2 \sin ^{2}\left(\frac{k h}{2}\right)}{(h) \sin (h)}\right\}=\frac{1}{2} \end{array}

As this limit can be evaluated directly by putting value of h because it is taking indeterminate form (0/0)
Thus, we use sandwich or squeeze theorem according to which -

\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \Rightarrow {\lim_{h\rightarrow 0} }\left\{\frac{\sin ^{2}\left(\frac{k h}{2}\right)}{(h) \sin (h)}\right\}=\frac{1}{4}

Dividing and multiplying by (kh/2)^2 to match the form in formula we have-

\begin{aligned} &\lim _{h \rightarrow 0}\left\{\frac{\sin ^{2}\left(\frac{\mathrm{kh}}{2}\right)}{(\mathrm{h}) \sin (\mathrm{h}) \times\left(\frac{\mathrm{kh}}{2}\right)^{2}} \times\left(\frac{\mathrm{kh}}{2}\right)^{2}\right\}=\frac{1}{4}\\ &\text { Using algebra of limits we get - }\\ &\lim _{h \rightarrow 0}\left(\frac{\sin \frac{k h}{2}}{\frac{k h}{2}}\right)^{2} \times \lim _{h \rightarrow 0} \frac{k^{2}}{4}\left(\frac{h}{\sin h}\right)=\frac{1}{4} \end{aligned}

\begin{aligned} &\text { Applying the formula- }\\ &\Rightarrow 1 \times\left(\mathrm{k}^{2} / 4\right)=(1 / 4)\\ &\Rightarrow \mathrm{k}^{2}=1\\ &\Rightarrow(k+1)(k-1)=0\\ &\therefore \mathrm{k}=1 \text { or } \mathrm{k}=-1 \end{aligned}

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infoexpert22

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