Trigonometric limit I couldn't find this limit , can someone help me?
$$\lim_{x \to \infty} \frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}$$ 

Hint. By using Taylor series expansions, as $u \to 0$, one has
$$
\begin{align}
\arctan u&=u-\frac{u^3}{3}+o(u^3)
\\\sin u&=u-\frac{u^3}{6}+o(u^3)
\end{align}
$$ giving, as $x \to \infty$ $\big($with $u=\frac1{x}$ or $u=\frac1{x+1}$$\big)$,
$$
\begin{align}
\arctan x&=\frac{\pi}2-\arctan \frac1x =\frac{\pi}2-\frac1x+O\left(\frac1{x^3} \right)
\\\arctan (x+1)&=\frac{\pi}2-\arctan \frac1{x+1} =\frac{\pi}2-\frac1x+\frac1{x^2} +O\left(\frac1{x^3} \right)
\\\sin \frac1x&=\frac1x+O\left(\frac1{x^3} \right)
\\\sin \frac1{x+1}&=\frac1x-\frac1{x^2} +O\left(\frac1{x^3} \right)
\end{align}
$$ then, as $x \to \infty$,

$$
\frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}=\frac{\frac1{x^2}+O\left(\frac1{x^3} \right)}{-\frac1{x^2}+O\left(\frac1{x^3} \right)} \to -1.
$$

