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.
 A: 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 $k$.
A: 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.
