Question

\[ f(x)=\left\{\begin{array}{ll} \frac{1-\cos k x}{x \sin x} & \text { if } x \neq 0 \\ 1 / 2 & \text { if } x=0 \end{array}\right. \] find \( K \) if \( x \) is continvous at \( x=0 \)

Answer 1

∵ f is continuous at x = 0.

Then f(0) = \(\underset{x\rightarrow0}{lim}\) f(x)

⇒ \(\underset{x\rightarrow0}{lim}\) \(\frac{1-cos\,kx}{x\,sin\,x}\) = 1/2

⇒ \(\underset{x\rightarrow0}{lim}\) \(\frac{k\,sin\,kx}{x\,cos\,x+sin\,x}\) = 1/2 (By D.L.H. Rule)

⇒ \(\underset{x\rightarrow0}{lim}\) \(\cfrac{k^2x\,\frac{sin\,kx}{kx}}{x(cos\,x+\frac{sin\,x}x)}\) = 1/2

⇒ \(\underset{x\rightarrow0}{lim}\) \(\cfrac{k^2\times\frac xx\times\frac{sin\,kx}{kx}}{cos\,x+\frac{sin\,x}x}\) = 1/2

⇒ k² \(\cfrac{\underset{kx\rightarrow0}{lim}\frac{sin\,kx}{kx}}{\underset{x\rightarrow0}{lim}\,cos\,x+\underset{x\rightarrow0}{lim}\,\frac{sin\,x}x}\) = 1/2

⇒ \(\frac{k^2\times1}{cos\,0+1}\) = 1/2 (∵ \(\underset{x\rightarrow0}{lim}\) sinx/x = 1)

⇒ k²/2 = 1/2

⇒ k² = 1

⇒ k = ±1

for k = 1 or k = -1, the given function f(x) is continuous.

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