Limits problem with trig? Factoring $\cos (A+B)$? Disclaimer: I am an adult learning Calculus. This is not a student posting his homework assignment. I think this is a great forum!
$$\lim_{h\to0} \frac{\cos(\frac{\pi}{3}+h)-\frac{1}{2}}{h}$$
Do I use the angle addition formula to do this?
I did that, and have no idea where to go from there.
$$\lim_{h\to0}\frac{\cos\frac{\pi}{3}\cos(h)-\sin\frac{\pi}{3}\sin(h)-\frac{1}{2}}{h}$$
What now?
 A: You might want to use the values of sine and cosine of $\pi/3$ (remember it is 60 degrees), and reduce the problem to evaluating the following two limits: $\lim \frac{1-\cos h}{h}$ and $\lim \frac{\sin h}{h}$. Then note that $1-\cos h$ is a higher order infinitesimal than $h$ so this term drops out. I will leave the second limit to you.
A: Let $\phi(x) = \cos(\frac{\pi}{3}+x)$. Note that $\phi(0) = \frac{1}{2}$, and $\phi$ is differentiable, with $\phi'(x) = -\sin(\frac{\pi}{3}+x)$.
So, the limit is $\lim_{h \to 0} \frac{\phi(h)-\phi(0)}{h} = \phi'(0) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$.
A: $$\frac{d(\cos x)}{dx}=\lim_{h\to0}\frac{\cos(x+h)-\cos x}h=\lim_{u\to0}\frac{\cos(x+2u)-\cos x}{2u}$$
Using $\displaystyle\cos C-\cos D=2\sin\frac{C+D}2\sin\frac{D-C}2$
$$\lim_{u\to0}\frac{\cos(x+2u)-\cos x}{2u}=-\frac12\cdot\lim_{u\to0}\frac{2\sin(x+u)\sin(-u)}{(-u)}=-\lim_{u\to0}\sin(x+u)\cdot\lim_{u\to0}\frac{\sin(-u)}{(-u)}$$
We know $\displaystyle\lim_{x\to0}\frac{\sin x}x=1$
Now in the current context $x=\frac\pi3$ as $\frac12=\cos\frac\pi3,$
Can you take it from here?
A: In this case, we can also use L'Hospital Rule because $\lim_{h \to 0} h =0$ and $\lim_{h \to 0} (cos(\pi/3+h)-1/2) =(cos(\pi/3+0)-1/2)=0$. $$\lim_{h \to 0}\frac{cos(\pi/3+h)-1/2}{h} =\frac{\lim_{h \to 0}(\frac{d}{dh}(cos(\pi/3+h)-1/2))}{\lim_{h \to 0}(\frac{d}{dh}h)}=\frac{\lim_{h \to 0}(-sin(\pi/3+h))}{\lim_{h \to 0}1}=\frac{-sin(\pi/3+0)}{1}=-\sqrt{3}/2.$$
