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

Oct 26, 2016

Use L'Hôpital's rule lim_(x→pi)sin(x)/(x - pi)= lim_(x→pi)cos(x) = -1

#### Explanation:

The derivative of the numerator is $\cos \left(x\right)$

The derivative of the denominator is 1 so we will not write it.

lim_(x→pi)sin(x)/(x - pi) = lim_(x→pi)cos(x) = -1

NOTE
The question was posted in "Determining Limits Algebraically" , so the use of L'Hôpital's rule is NOT a suitable method to solve the problem

Oct 27, 2016

I see that substitution gets us the indeterminate form $\frac{0}{0}$.

#### Explanation:

A limit like this reminds me that ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$

And that makes me wonder if can use $\sin \left(x - \pi\right)$ somehow.

The difference formula for the sine gets us:

$\sin \left(x - \pi\right) = \sin x \cos \pi - \cos x \sin \pi = \sin x \left(- 1\right) - \cos x \left(0\right) = - \sin x$

Therefore we can replace $\sin x$ by $- \sin \left(x - \pi\right)$

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

$= - {\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta}$ $\text{ }$ (with $\theta = x - \pi$)

$= - 1$