Non-circular proof of $\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$

Solution 1:

Usually the easiest to understand geometric justification is via the area of the sector of the unit disk associated with $θ$. What should be clear by the geometric definition of the angle is that this area is proportional to $θ$, $A(θ)=Cθ$.

Using triangles inside the sector you get a lower bound of either $\frac12\cosθ\sinθ=\frac14\sin2θ$ or $\sin\fracθ2\cos\fracθ2=\frac12\sinθ$. Using triangles outside the unit disk, one gets upper bounds of either $\frac12\tan θ$ or $\tan \fracθ2=\frac{\sinθ}{1+\cosθ}$. Thus whatever the measure for the angle is, you get $$ \frac12\sinθ\le Cθ\le \frac{\sinθ}{1+\cosθ}\iff C·(1+\cosθ)\le\frac{\sinθ}{θ}\le 2C $$ which results in $\lim_{θ\to0}\frac{\sinθ}{θ}=2C$.

What value to assign to $C$ and closely related what measurement to assign to the angle of the full circle has now entirely analytical reasons to keep certain series from being cluttered with arbitrary constants. Without that it is completely reasonable to assign $θ=360°$ to the full circle and have $C=\frac\pi{360°}$.

Solution 2:

$$\sin t=\frac{e^{it}-e^{-it}}{2i}.$$

Hence,

$$ \frac{\sin t}{t}= \frac{1+it+\epsilon_1(t)-1+it+\epsilon_2(t)}{2it}=\frac{2it+\epsilon(t)}{2it}=1+\frac{\epsilon(t)}{2it}, $$

where $\epsilon(t)/2it \rightarrow 0$ as $t \rightarrow 0$.

There is no circularity: We don't "use the derivatives of $\sin$ to arrive at the Taylor series". It is the definition (one of the possible ones, but this is one good to work with). Motivating the definition is another issue altogether.

For instance, you could prove that there exists at most one pair of functions $s,c$ such that $s'=c$, $c'=-s$, $s^2+c^2\equiv 1$ and $s(0)=0$. These properties are easy to "prove" for the case where $\sin$ is "geometrically" defined (the definition of this is by drawing... so the best you can get is arguments by drawing too. It is not circular, it is just ill-founded) and easy to prove from the definition above. Hence, they correspond.