1. Tardigrade
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  3. Mathematics
  4. Let φ(x)=(1-cosλ/x sin x), x≠ 0 and φ(0(1/2) . If φ(x) is continuous at x = 0, then λ is

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Q. Let $\phi(x)=\frac{1-cos\lambda}{x sin x}, x\neq 0$ and $\phi(0\frac{1}{2}$ . If $\phi(x) $ is continuous at $x$ = 0, then $\lambda $ is

Solution:

$\lim_{x\to0} \phi \left(x\right) = \lim _{x\to 0} \frac{1 - \cos \lambda x}{x \sin x} = \lim _{x\to 0} \frac{ 1 - \cos \lambda x}{x^{2}} $
= $ \lim _{x\to 0} \frac{2 sin^{2} \frac{\lambda x}{2}}{x^{2}} $
= $\lim _{x\to 0} 2\left(\frac{\frac{\sin \frac{\lambda x}{2}}{\lambda x}}{2}\right)^2 \frac{\lambda^{2}}{4} = \frac{\lambda^{2}}{2}$
Also $\phi ( 0) = \frac{1}{2}$
Since $\phi (x)$ is continuous at $x$ = 0
$\therefore $ $\lim_{x\to0} \phi \left(x\right) = \phi (0)$
$\Rightarrow \, \frac{\lambda^2}{2} = \frac{1}{2} \Rightarrow \lambda^2 = 1 \Rightarrow \lambda = \pm 1 $