# How do you find the limit of sec3xcos5x as x approaches pi/2 from the left?

Aug 20, 2016

Reqd. Limit$= - \frac{5}{3}$.

#### Explanation:

Reqd. Limit$= {\lim}_{x \rightarrow \frac{\pi}{2} -} \sec 3 x \cos 5 x$

Since $x \rightarrow \frac{\pi}{2} - , x < , \frac{\pi}{2}$. Let $x + h = \frac{\pi}{2}$

$\therefore x \rightarrow \frac{\pi}{2} - \Rightarrow h \rightarrow 0 +$

Now, $\sec 3 x \cos 5 x = \sec 3 \left(\frac{\pi}{2} - h\right) \cos 5 \left(\frac{\pi}{2} - h\right)$

$= \sec \left(3 \frac{\pi}{2} - 3 h\right) \cos \left(5 \frac{\pi}{2} - 5 h\right) = \left(- \csc 3 h\right) \left(\sin 5 h\right) = - \sin \frac{5 h}{\sin} \left(3 h\right)$

$= - \left[\frac{\left(\sin \frac{5 h}{5 h}\right) \cdot 5 h}{\left(\sin \frac{3 h}{3 h}\right) \cdot 3 h}\right]$

$= - \frac{5}{3} \left[\frac{\left(\sin \frac{5 h}{5 h}\right)}{\left(\sin \frac{3 h}{3 h}\right)}\right]$

Since, ${\lim}_{h \rightarrow 0 +} \sin \frac{5 h}{5 h} = {\lim}_{h \rightarrow 0 +} \sin \frac{3 h}{3 h} = 1$,

Reqd. Limit$= - \frac{5}{3}$.

Enjoy Maths.!