Answer to Question #329132 in Calculus for Himanshi

Question #329132

The following function is continuous at x = 0:


𝑓(𝑥) =


(1−cos𝑘𝑥)/(xtanx)


, 𝑓𝑜𝑟 𝑥 ≠ 0 𝑎𝑛𝑑 𝑓(0) = 3. find k.

1
Expert's answer
2022-04-17T10:08:40-0400

Using the well-known trigonometric identity, we get: 1coskx=2sin2k2x1-\cos\,kx=2\,sin^2\frac{k}{2}x. Thus, the function can be rewritten as: 2sin2k2xxtanx\frac{2\sin^2\frac{k}{2}x}{x\,\tan\,x}. Remind the well-known limit: sinxx1,x0\frac{\sin\,x}{x}\rightarrow1,x\rightarrow0. We rewrite the expression and receive: xk242sin2k2xk24x2tanx\frac{xk^2}{4}\frac{2\sin^2\frac{k}{2}x}{\frac{k^2}{4}x^2\,\tan\,x}. Since, xtanx1,x0\frac{x}{\tan\,x}\rightarrow1,x\rightarrow0 and sinxx1,x0\frac{\sin\,x}{x}\rightarrow1,x\rightarrow0, we get: xk242sin2k2xk24x2tanxk24,x0\frac{xk^2}{4}\frac{2\sin^2\frac{k}{2}x}{\frac{k^2}{4}x^2\,\tan\,x}\rightarrow\frac{k^2}{4},x\rightarrow0. From the latter and from the formulation of the task we get: k24=3\frac{k^2}{4}=3. Finally, we get: k=±12.k=\pm\sqrt{12}. Thus, the answer is: k=±12.k=\pm\sqrt{12}.


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