Limit as $y\to x$ of $(\sin y-\sin x)/(y-x)$ without L’Hospital

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 $\cos x$.

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 $\cos x$.