# How do you find the limit cosx/(pi/2-x) as x->pi/2?

Feb 13, 2017

Because the expression evaluated at the limit results in an indeterminate form (specifically $\frac{0}{0}$), one should use L'Hôpital's rule .

#### Explanation:

Given: lim_(xto"pi/2)cos(x)/(pi/2-x)=?

Using L'Hôpital's rule , we compute the derivative of the numerator and the denominator:

$\frac{d \left(\cos \left(x\right)\right)}{\mathrm{dx}} = - \sin \left(x\right)$

$\frac{d \left(\frac{\pi}{2} - x\right)}{\mathrm{dx}} = - 1$

Assemble the new expression and evaluate at the limit:

lim_(xto"pi/2)(-sin(x))/(-1)=1

According to the rule, the original limit goes to the same value:

lim_(xto"pi/2)cos(x)/(pi/2-x)=1

Feb 13, 2017

If you have not yet learned l'Hospital's Rule, then see below.

#### Explanation:

Use ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$ after using

$\cos x = \sin \left(\frac{\pi}{2} - x\right)$ (co-functiono identity from trigonometry).

${\lim}_{x \rightarrow \frac{\pi}{s}} \cos \frac{x}{\frac{\pi}{2} - x} = {\lim}_{x \rightarrow \frac{\pi}{2}} \sin \frac{\frac{\pi}{2} - x}{\frac{\pi}{2} - x}$

Note that: $\lim \left(x \rightarrow \frac{\pi}{2}\right) \left(\frac{\pi}{2} - x\right) = 0$, so with $\theta = \frac{\pi}{2} - x$, we have

$= {\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$