Limit as $y\to x$ of $(\sin y-\sin x)/(y-x)$ without L’Hospital $$\lim_{y\rightarrow x}\frac{\sin(y)-\sin(x)}{y-x}$$
Is there any cool trig identity I could use to solve this? We don't have L’Hospital yet, so I have to calculate it otherwise. I tried solving this using the series expansion of sine:
$$\cdots =\lim_{y\rightarrow x}\frac{\left(\sum_{k=0}^\infty (-1)^k \dfrac{y^{2k+1}}{(2k+1)!}\right) -\left(\sum_{k=0}^\infty (-1)^k \dfrac{x^{2k+1}}{(2k+1)!}\right)}{y-x}$$
But what now? With L’Hospital I get $\cos(x)$ as a solution. Differentiation isn't allowed either.
 A: Render
$\sin y -\sin x= 2\sin(\frac{y-x}{2})\cos(\frac{y+x}{2})$
Assume you know that $g(u)=(\sin u)/u$ has the limit $1$ as $u\to 0$.  Then we have, using the above result:
$\dfrac{\sin y -\sin x}{y-x}= 2\dfrac{g(\frac{y-x}{2})(y-x)}{2(y-x)}\cos(\frac{y+x}{2})$
and the zero factor in the denominator is now cancelled out leading to the overall limit $\cos x$.
A: Note that
\begin{align*}
\dfrac{1}{y-x}(y^{2k+1}-x^{2k+1})=y^{2k}+y^{2k-1}x+\cdots+yx^{2k-1}+x^{2k}\rightarrow(2k+1)x^{2k}.
\end{align*}
So you end up with $\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^{k}}{(2k)!}x^{2k}$, which is $\cos x$.
A: In the comments, someone pointed out that you can show that $\lim_{y\to x}\dfrac{\sin(y)-\sin(x)}{y-x}=\cos(x)$ if we know that:
$$\lim_{\theta\to0}\dfrac{\sin(\theta)}{\theta}=1,\quad(1)$$
$$\lim_{\theta\to0}\dfrac{\cos(\theta)-1}{\theta}=0.\quad(2)$$
Here is a link to that argument: Solving a limit given a limit
I will discuss how we can show the above limits without using L'Hôpital's rule.
To prove (1), first we use some geometry to prove that for any $\theta$ with $0<\theta<\frac{\pi}{2}$, we have that
$$0<\cos(\theta)<\dfrac{\sin(\theta)}{\theta}<\dfrac{1}{\cos(\theta)}.\quad(3)$$
I will discuss how to prove this below. But first let me show how we can use this to evaluate the limits above. Note that the three functions in (3) above are all even. So the above inequality holds for all non-zero $\theta$ with $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$. It follows from the squeeze theorem that:
$$\lim_{\theta\to0}\dfrac{1}{\cos(\theta)}=\lim_{\theta\to0}\dfrac{\sin(\theta)}{\theta}=\lim_{\theta\to0}\cos(\theta)=1.$$
To prove (2), note that
$\begin{align*}
\lim_{\theta\to0}\dfrac{\cos(\theta)-1}{\theta} &=\lim_{\theta\to0}\dfrac{(\cos(\theta)-1)}{\theta}\cdot\dfrac{(\cos(\theta)+1)}{(\cos(\theta)+1)} \\
&=\lim_{\theta\to0}\dfrac{\cos^2(\theta)-1}{\theta\cdot(\cos(\theta)+1)} \\
&=\lim_{\theta\to0}\dfrac{-\sin^2(\theta)}{\theta\cdot(\cos(\theta)+1)} \\
&=\lim_{\theta\to0}-\sin(\theta)\cdot\dfrac{\sin(\theta)}{\theta}\cdot\dfrac{1}{(\cos(\theta)+1)} \\
&=0.
\end{align*}$
To prove (3), let $0<\theta<\frac{\pi}{2}$, and let $A=(0,0)$, let $B=(\cos(\theta),0)$, let $C=(\cos(\theta),\sin(\theta))$, let $X=(1,0)$ and let $Y=(1,\tan(\theta))$. I apologize for not having a picture to go along with these definitions, but you can probably find this exact picture in any calculus textbook.
Now comparing the areas of triangle $ABC$, sector $AXC$, and triangle $AXY$, we have that
$$0<\frac{1}{2}\cos(\theta)\sin(\theta)<\frac{1}{2}\theta<\frac{1}{2}\tan(\theta).$$
Inverting this inequality gives
$$0<\dfrac{2\cos(\theta)}{\sin(\theta)}<\dfrac{2}{\theta}<\dfrac{2}{\cos(\theta)\sin(\theta)}.$$
If we multiply these inequalities by $\frac{1}{2}\sin(\theta)$ then we obtain (3).
