How do you find the limit of (1-sintheta)/(theta-pi/2) as theta->pi/2?

Feb 28, 2017

$0.$

Explanation:

Recall that, ${\lim}_{x \to a} \frac{f \left(x\right) - f \left(a\right)}{x - a} = f ' \left(a\right) .$

So, we have, ${\lim}_{\theta \to \frac{\pi}{2}} \frac{1 - \sin \theta}{\theta - \frac{\pi}{2}}$

$= - {\lim}_{\theta \to \frac{\pi}{2}} \frac{\sin \theta - \sin \left(\frac{\pi}{2}\right)}{\theta - \frac{\pi}{2}}$

$= - \sin ' \left(\frac{\pi}{2}\right)$

$= - \cos \left(\frac{\pi}{2}\right)$

$= 0.$

In the following Direct Method to find the limit, we will use the

Standard Limit : ${\lim}_{x \to 0} \sin \frac{x}{x} = 1.$

Let, $l = {\lim}_{\theta \to \frac{\pi}{2}} \frac{1 - \sin \theta}{\theta - \frac{\pi}{2}} .$

Subst. $\theta = h + \frac{\pi}{2} , \text{ so that, as } \theta \to \frac{\pi}{2} , h \to 0.$

$\therefore l = {\lim}_{h \to 0} \frac{1 - \sin \left(\frac{\pi}{2} + h\right)}{h} ,$

$= \lim \frac{1 - \cos h}{h} = \lim \left\{\frac{1 - \cos h}{h}\right\} \left\{\frac{1 + \cos h}{1 + \cos h}\right\}$

$= \lim {\sin}^{2} \frac{h}{h \left(1 + \cos h\right)}$

$= {\lim}_{h \to 0} \left\{\sin \frac{h}{h}\right\} \left\{\frac{\sinh}{1 + \cos h}\right\}$

$= \left(1\right) \left\{\frac{0}{1 + 1}\right\}$

$= 0.$

Enjoy Maths.!