limx->0 cosx-1/2x
ok i kinda dont know where to start
Is this $\displaystyle \displaystyle \lim_{x \to 0}\frac{\cos{x} - 1}{2x}$? If so
$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\cos{x} - 1}{2x} &= \frac{1}{2}\lim_{x \to 0}\frac{\cos{x} - 1}{x} \\ &= \frac{1}{2}\lim_{x \to 0}\frac{(\cos{x} - 1)(\cos{x} + 1)}{x(\cos{x} + 1)} \\ &= \frac{1}{2}\lim_{x \to 0}\frac{\cos^2{x} - 1}{x(\cos{x} + 1)} \\ &= \frac{1}{2}\lim_{x \to 0}\frac{-\sin^2{x}}{x(\cos{x} + 1)} \\ &= -\frac{1}{2}\lim_{x \to 0}\frac{\sin{x}}{x} \cdot \lim_{x \to 0}\frac{\sin{x}}{\cos{x} + 1}\\ &= -\frac{1}{2}\cdot 1 \cdot \frac{0}{2} \\ &= 0 \end{align*}$
Normally, I complain about people using L'Hopital's rule when simpler methods will work but I think that is the best way to handle this one:
$\displaystyle \lim_{x\to 0}\frac{cos(x)-1}{2x}= \lim_{x\to 0}\frac{-sin(x)}{2}= 0$
Or use a power series: $\displaystyle cos(1)= 1- (1/2)x^2+ (1/24)x^4+ \cdot\cdot\cdot$ so
$\displaystyle \frac{cos(x)-1}{2x}= \frac{-(1/2)x^2+ (1/24)x^4+ \cdot\cdot\cdot}{2x}= -(1/4)x+ (1/48)x^3+ \cdot\cdot\cdot$
which goes to 0 as x goes to 0.
Alternatively, one could just use the definition of the derivative.
$\displaystyle \lim_{x \to 0} \frac{\cos(x) - 1}{x} = \left. \frac d {dx} \cos(x) \right|_{x = 0} = -\sin(0) = 0.$
I guess this is probably begging the question a little bit, though, since we effectively need to know this limit to calculate the derivatives of sin and cos in the first place. If you already have $\displaystyle \frac{\sin(x)}{x} \to 1$ as $\displaystyle x \to 0$ and you are working with the geometric definitions (not the power series definitions), then Prove It's proof seems best so far.