Answer to Question #329132 in Calculus for Himanshi

Question

The following function is continuous at x = 0:

𝑓(𝑥) =

(1−cos𝑘𝑥)/(xtanx)

, 𝑓𝑜𝑟 𝑥 ≠ 0 𝑎𝑛𝑑 𝑓(0) = 3. find k.

Answer 1

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