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

Dec 5, 2016

Rewrite so we can use ${\lim}_{\theta \rightarrow 0} \frac{\theta}{\sin} \theta = 1$

Explanation:

$\cos \left(\frac{\pi}{2} - x\right) = \sin x$ (by co-function identity or difference formula for cosine)

Also $\cos \left(- \theta\right) = \cos \theta$ (cosine is an even function)

So $\cos \left(x - \frac{\pi}{2}\right) = \cos \left(\frac{\pi}{2} - x\right) = \sin x$ and

sinx = cos(x-pi/2)

$\sin \left(\frac{\pi}{2} - x\right) = \cos x$ (by co-function identity or difference formula for sine)

Also $\sin \left(- \theta\right) = - \sin \theta$ (sine is an odd function)

So $\sin \left(x - \frac{\pi}{2}\right) = - \sin \left(\frac{\pi}{2} - x\right) = - \cos x$and

cosx = -sin(x-pi/2)

Now we can rewrite

${\lim}_{x \rightarrow \frac{\pi}{2}} \left(\frac{\pi}{2} - x\right) \tan x = {\lim}_{x \rightarrow \frac{\pi}{2}} \left(\frac{\pi}{2} - x\right) \sin \frac{x}{\cos} x$

$= {\lim}_{x \rightarrow \frac{\pi}{2}} \left(\frac{\pi}{2} - x\right) \cos \frac{\frac{\pi}{2} - x}{- \sin \left(\frac{\pi}{2} - x\right)}$

$= {\lim}_{x \rightarrow \frac{\pi}{2}} - \frac{\frac{\pi}{2} - x}{\sin \left(\frac{\pi}{2} - x\right)} \cdot \cos \left(\frac{\pi}{2} - x\right)$

Notice that ${\lim}_{x \rightarrow \frac{\pi}{2}} \frac{\frac{\pi}{2} - x}{\sin \left(\frac{\pi}{2} - x\right)}$ is a version of ${\lim}_{\theta \rightarrow 0} \frac{\theta}{\sin} \theta$, so we get

$= - \left(1\right) \left(1\right) = - 1$