# Prove that $\lim_{x\to0}\frac{\sin x}x=1$ using algebraic manipulations of derivatives?

I can prove that $\lim_{x\to0}\frac{\cos(x)-1}x=0$ since

$$\sin^2(x)=1-\cos^2(x)$$

$$\implies2\sin'(x)\sin(x)=-2\cos'(x)\cos(x)$$

$$\sin'(x)\sin(x)=-\cos'(x)\cos(x)$$

at $x=0$, we have

$$0=-\cos'(0)$$

Thus, $\cos'(0)=\lim_{x\to0}\frac{\cos(x)-1}x=0$.

Can one produce the same result for the famous $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ by manipulating derivatives?

Particularly, can we calculate $\sin'(0)$ without first showing that $\sin'(x)=\cos(x)$?

Edit:

As has been shown, we need more than just trig identities to prove this, since trig identities work regardless of the radian/degrees while the limit does not. So consider the following information:

$$0\le\frac{\sin(x+t)-\sin(x)}t\le\cos(x)\ \forall\ x\in(0,\frac\pi2),\ t\in\left(0,\frac\pi2-x\right)$$

The last inequality proven geometrically in this answer.

Thus, we get

$$0\le\sin'(0)\le\cos(0)$$

As of yet, I'm unsure what other information should be required, mainly how to deal with the units issue.

• How do you know that $\sin^{\prime}=\cos$ ? Usually this is derived as a consequence of the limit in your title. – Rene Schipperus Dec 30 '16 at 0:27
• @SimpleArt How would you know from the graph that nothing funny happens around $x=10^{-\text{Graham's number}}$? – Arthur Dec 30 '16 at 0:33
• @ReneSchipperus From looking at the question, I'd say it's pretty clear he's assumed that 1) $\sin'$ exists and is bounded near $0$, and that 2) normal differentiation rules, specifically the chain rule, applies. – Arthur Dec 30 '16 at 0:36
• If all you are given is that $s(x)$ and $c(x)$ are functions which satisfy $s(x)^2+c(x)^2 = 1$ for all real $x$ and $s(0) = 0$ and $c(0) = 1$, then both [$s(x) = \sin(x)$, $c(x) = \cos(x)$] and [$s(x) = -\sin(x)$, $c(x) = \cos(x)$] are solutions. So it won't be possible to decide whether $s'(0) = 1$ or $s'(0) = -1$. – JimmyK4542 Dec 30 '16 at 1:07
• @SimpleArt: The point of my comment was that using only the information $\sin^2(x)+\cos^2(x) = 1$, $\sin(0) = 0$, $\cos(0) = 1$ and manipulating derivatives will not be enough to determine $\sin'(0)$. If you have other assumptions about $\sin(x)$ and $\cos(x)$, then you might be able to prove $\sin'(0) = 1$. – JimmyK4542 Dec 30 '16 at 1:25

I don't think it is possible to derive that $\sin'(0)=1$ using only "trigonometric identities", due to the fact that trigonometric identities (at least what I call trigonometric identities - see (*)) are blind with respect to the essential fact that makes $\frac{\sin(x)}{x} \to 1$, which is, informally speaking, the measurement by "radians".

What we can do using only trigonometric identities is derive the fact that $\sin'(x)=\sin'(0) \cos(x)$ and $\cos'(x)=-\sin'(0)\sin(x)$. For the first, consider the identity $$\sin(x+y)=\sin(x)\cos(y) +\sin(y)\cos(x).$$ Differentiating with respect to $y$, we get $$\sin'(x+y)=\sin(x)\cos'(y)+\sin'(y)\cos(x).$$ Evaluating at $y=0$, $$\sin'(x)=\sin(x)\cos'(0)+\sin'(0)\cos(x)$$ $$\therefore \sin'(x)=\sin'(0)\cos(x),$$ since you concluded that $\cos'(0)=0$. Analogously, using the identity for $\cos(x+y)$, one reaches the other formula.

(*) The question is quite unclear. Using only "trigonometry", we are left with a fair amount of freedom on the functions $\sin$, $\cos$ as real functions (essentially, changing $\sin(x)$ to $\sin(kx)$ for some constant $k \neq 0$ does not change trigonometry, which is what we perceive in practice as a "change of units" on the angles. And this amounts to changing $\sin'(0)$ as well, by the same factor). More explicitly with respect to the question, I consider a "trigonometric identity" to entail (not iff) that it is invariant under changing the functions $\sin(x)$, $\cos(x)$ by $\sin(k x)$, $\cos(kx)$. As such, it is thus impossible to prove that $\sin'(0)=1$ using only "trigonometric identities", because there is always a factor on the derivative which can come from the constant $k$.

• Hm, you make very good points at the end. It comes to be that regardless of units, I can derive $\cos'(0)=0$, but not for $\sin'(0)$. I'll think about that if I can figure out any ways to improve the question. – Simply Beautiful Art Dec 30 '16 at 1:35

We can easily see that the trigonometric relations

$$\sin (0)=0, \cos (0)=1, \sin ^2(x)+\cos ^2(x)=1$$

are not sufficient to prove

$$\lim_{x\to0}{\sin(x)\over x} = 1$$

Indeed, define two functions

$$\text{sen}(x)=\sin (a x), \text{ces}(x)=\cos (a x)$$

These satisfy the conditions above but

$$\lim_{x\to 0} \, \frac{\text{sen}(x)}{x}=a$$

which obviously need not be unity.

• No worries about $\LaTeX$, it won't stop me from fixing it up :D – Simply Beautiful Art Dec 30 '16 at 1:48