# Rigor in showing limit of $\frac{\cos(\Delta x)-1}{\Delta x}=0$ [duplicate]

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?

• Are you ok by saying that $\cos(x) = 1 - x^2/2 + o(x^2)$ when $x$ is close to $0$ ? – Zubzub Sep 27 '18 at 13:41
• Since we're trying to establish calculus, can you show that your approximation works without using calculus to derive the $o(x^2)$ approximation? – spraff Sep 27 '18 at 13:44

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

• I don't see how. You still haven't established that the top approaches zero faster than the bottom. – spraff Sep 27 '18 at 13:42
• lim sinx / x = 1 – Mikey Spivak Sep 27 '18 at 13:43
• I know, but you haven't established that the top approaches zero faster than the bottom. – spraff Sep 27 '18 at 13:44
• The limits $\frac{\sin x}x$ and $\frac{\sin x}{1+\cos x}\$ exist as $x\to 0$. So we are indeed done – user418131 Sep 27 '18 at 13:46

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

$$\frac{\cos x - 1 }{x}=x\cdot \frac{\cos x - 1 }{x^2}\to 0$$

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 $$\left|\sin x\over x\right|\le1$$.

• +1 I had written almost similar answer (now deleted). – Paramanand Singh Sep 27 '18 at 15:53

$$\frac{\cos\Delta x-1}{\Delta x}=-2\frac{\sin^2\dfrac{\Delta x}2}{\Delta x}=-\sin\dfrac{\Delta x}2\frac{\sin\dfrac{\Delta x}2}{\dfrac{\Delta x}2}.$$

Now it is enough to show that the last fraction tends to a finite limit.