# Can someone help me find a and b if f(x) is continuous at π/2 , without using L'hospital rule? Please mention all formulae used too.

## Jun 12, 2018

$a = \frac{1}{2} , b = 4$

#### Explanation:

To find the left hand limit ${\lim}_{x \to {\left(\frac{\pi}{2}\right)}^{-}} f \left(x\right)$ we note that for $x < \frac{\pi}{2}$

$f \left(x\right) = \frac{1 - {\sin}^{3} x}{3 {\cos}^{2} x}$
$= \frac{\left(1 - \sin x\right) \left(1 + \sin x + {\sin}^{2} x\right)}{3 \left(1 - \sin x\right) \left(1 + \sin x\right)}$
$= \frac{1 + \sin x + {\sin}^{2} x}{3 \left(1 + \sin x\right)}$

where the last step is legitimate as one approaches $\frac{\pi}{2}$ because $\left(1 - \sin x\right)$ is non-zero as long as $x \ne \frac{\pi}{2}$

Thus

${\lim}_{x \to {\left(\frac{\pi}{2}\right)}^{-}} f \left(x\right) = {\lim}_{x \to \frac{\pi}{2}} \frac{1 + \sin x + {\sin}^{2} x}{3 \left(1 + \sin x\right)} = \frac{1}{2}$

Hence for continuity, we have

$\textcolor{red}{a = \frac{1}{2}}$

To figure out the right hand limit we use

$\frac{b \left(1 - \sin x\right)}{\pi - 2 x} ^ 2$
$= \frac{b}{16} \frac{{\sin}^{2} \left(\frac{x}{2}\right) + {\cos}^{2} \left(\frac{x}{2}\right) - 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{\frac{x}{2} - \frac{\pi}{4}} ^ 2$
$= \frac{b}{16} {\left(\sin \left(\frac{x}{2}\right) - \cos \left(\frac{x}{2}\right)\right)}^{2} / {\left(\frac{x}{2} - \frac{\pi}{4}\right)}^{2}$
$= \frac{b}{8} {\left(\sin \left(\frac{x}{2}\right) \frac{1}{\sqrt{2}} - \cos \left(\frac{x}{2}\right) \frac{1}{\sqrt{2}}\right)}^{2} / {\left(\frac{x}{2} - \frac{\pi}{4}\right)}^{2}$
$= \frac{b}{8} {\left(\sin \left(\frac{x}{2}\right) \cos \left(\frac{\pi}{4}\right) - \cos \left(\frac{x}{2}\right) \sin \left(\frac{\pi}{4}\right)\right)}^{2} / {\left(\frac{x}{2} - \frac{\pi}{4}\right)}^{2}$
$= \frac{b}{8} {\sin}^{2} \frac{\frac{x}{2} - \frac{\pi}{4}}{\frac{x}{2} - \frac{\pi}{4}} ^ 2$

Hence the right hand limit is

${\lim}_{x \to {\left(\frac{\pi}{2}\right)}^{+}} f \left(x\right) = {\lim}_{x \to \left(\frac{\pi}{2}\right)} \frac{b}{8} {\sin}^{2} \frac{\frac{x}{2} - \frac{\pi}{4}}{\frac{x}{2} - \frac{\pi}{4}} ^ 2$
$q \quad = \frac{b}{8} {\lim}_{\left(\frac{x}{2} - \frac{\pi}{4}\right) \to 0} {\left(\sin \frac{\frac{x}{2} - \frac{\pi}{4}}{\frac{x}{2} - \frac{\pi}{4}}\right)}^{2} = \frac{b}{8}$

Thus for $f \left(x\right)$ to be continuous, we must also have

$\frac{b}{8} = \frac{1}{2} \implies \textcolor{red}{b = 4}$