What is $ \lim_{x\to\pi} \frac {1+\cos{x}}{{x-\pi}}$? How do I solve it? Is there some crafty algebraic step I need to follow? I'm guessing it's $\pi$ in the end..
$$ \lim_{x\to\pi} \frac {1+\cos{x}}{{x-\pi}}$$
Note: L'Hopital's rule and derivatives are not allowed
 A: Ok, No L'Hospital. Here is the set up: Set $x-\pi=t$ and the limit expression becomes $$ \lim_{t\to 0} \frac {1+\cos({t+\pi})}{t}$$ Now using $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$ you can rewrite numerator as $1+\cos(t)\cos(\pi)-\sin(t)\sin(\pi)$ and since $\sin(\pi)=0$, the new limit becomes: 
$$ \lim_{t\to 0} \frac {1-\cos(t)}{t}$$ This is of course still indeterminate. Multiply top and bottom by the conjugate of the numerator and resort to standard limits to finish the problem. Please give it a try from here...
Here is an alternative, rewrite the limit as $$ \lim_{x\to\pi} \frac {\cos{x}-\cos\pi}{{x-\pi}}$$ and use the limit definition of the derivative of cosine (if that is allowed by your instructor?)
A: You can do this without derivatives by using trig identities and the geometric inequality $\sin\theta\le\theta$ for $\theta\ge0$.  (By "geometric," I mean that $\sin\theta$ is the $y$ coordinate of the point $(\cos\theta,\sin\theta)$ on the unit circle, while $\theta$ is the arc length along the circle from $(1,0)$ to $(\cos\theta,\sin\theta)$.)  
Letting $\theta=x-\pi$, we have
$$\lim_{x\to\pi}{1+\cos x\over x-\pi}=\lim_{\theta\to0}{1+\cos(\theta+\pi)\over\theta}=\lim_{\theta\to0}{1-\cos\theta\over\theta}=\lim_{\theta\to0}{1-\cos^2\theta\over\theta(1+\cos\theta)}={1\over2}\lim_{\theta\to0}{\sin^2\theta\over\theta}$$
Utilizing the geometric inequality, we have
$$0\le\left|\sin^2\theta\over\theta\right|={\sin^2|\theta|\over|\theta|}\le{|\theta|^2\over|\theta|}=|\theta|\to0$$
Hence by the Squeeze Theorem the limit is $0$.
A: With some trig identities:
$$\lim_{x\to \pi} \frac{1+\cos x}{x-\pi} = \lim_{x\to \pi} \frac{\cos x- \cos\pi}{x-\pi}= \lim_{x\to \pi} \frac{-2\sin((x+\pi)/2)\sin((x-\pi)/2)}{x-\pi}$$
$$=\lim_{x\to \pi}\left(\frac{-2\sin((x+\pi)/2)}{2}\right)\left(\frac{\sin((x-\pi)/2)}{(x-\pi)/2}\right)  = \lim_{x\to \pi}-\sin((x+\pi)/2)\cdot 1 = 0.$ $$
A: Write $x-\pi=2t$
$$ \lim_{x\to\pi} \frac {1+\cos{x}}{{x-\pi}}=\lim_{t\to0}\dfrac{1+\cos(\pi+2t)}{2t}=\lim_{t\to0}\dfrac{1-\cos2t}{2t}$$
$$=\left(\lim_{t\to0}\dfrac{\sin t}t\right)^2\cdot(\lim_{t\to0}t)=?$$
