Rigor in showing limit of $\frac{\cos(\Delta x)-1}{\Delta x}=0$ The standard derivation for $\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)$ involves using the fact that
$$\lim_{\Delta x \to 0}\frac{\cos(\Delta x)-1}{\Delta x}=0$$
It's trivial to show that
$$\lim_{\theta\to 0} \cos(\theta)-1= 0$$
but the first limit doesn't follow directly from the second without a bit more work to handle the division.
How can we make this more rigorous?
 A: We know that
$$\frac{\cos x - 1 }{x^2}= \frac{\cos x - 1 }{x^2}\frac{\cos x + 1 }{\cos x + 1}=-\frac{\sin^2 x }{x^2}\frac{1 }{\cos x + 1}\to -\frac12$$
and therefore
$$\frac{\cos x - 1 }{x}=x\cdot \frac{\cos x - 1 }{x^2}\to 0$$
A: From the geometric definition of $\sin\theta$ as the $y$ coordinate of the point on the unit circle at angle $\theta$ with respect to the $x$ axis, we have $|\sin\theta|\le|\theta|$ for all angles $\theta$. Using the trigonometric identity $\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)$, it follows, for $|x|\le\pi/2$ (so that $1+\cos x\ge1$), that
$$0\le\left|\cos x-1\over x\right|=\left|-\sin^2x\over x(\cos x+1) \right|=|\sin x|\left|\sin x\over x\right|\left|1\over1+\cos x\right|\le|\sin x|\cdot1\cdot1=|\sin x|\to0\quad\text{as }x\to0$$
and thus
$$\lim_{x\to0}\left(\cos x-1\over x\right)=0$$
by the Squeeze Theorem.  Note, in particular, that it is not necessary to know that ${\sin x\over x}\to1$ as $x\to0$, it suffices to know that $\left|\sin x\over x\right|\le1$.
A: notice that 
$$ \frac{\cos x - 1 }{x} \cdot \frac{\cos x + 1 }{\cos x + 1 } = \frac{\cos^2 x - 1}{x(\cos x + 1 )} = - \frac{ \sin x \cdot \sin x }{x (\cos x + 1)}$$
The rest is history
A: $$\frac{\cos\Delta x-1}{\Delta x}=-2\frac{\sin^2\dfrac{\Delta x}2}{\Delta x}=-\sin\dfrac{\Delta x}2\frac{\sin\dfrac{\Delta x}2}{\dfrac{\Delta x}2}.$$
Now it is enough to show that the last fraction tends to a finite limit.
