Trigonometric limit I couldn't find this limit , can someone help me?
$$\lim_{x \to \infty} \frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}$$ 
 A: Use identities:
$$
\sin\theta -\sin\varphi=2\sin\left(\theta-\varphi\over2\right)\cos\left(\theta+\varphi\over2\right)\\
\arctan x-\arctan y=\arctan\left(x-y\over1+xy\right)
$$
Then,
$$
\lim_{x \to \infty} \frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}=
\lim_{x \to \infty} \frac{\arctan\left(\frac{1}{1+x(x+1)}\right)}{2\sin\left(\frac{1}{x+1}-\frac{1}{x}\over2\right)\cos\left( \frac{1}{x+1}+\frac{1}{x}\over2\right)}\to\\
\lim_{x \to \infty}\frac{\arctan\left(\frac{1}{1+x+x^2}\right)}{2\sin\left(-\frac{1}{2x(x+1)}\right)\cos\left(\frac{2x+1}{2x(x+1)}\right)}\to-\frac{1}{2}\lim_{x \to \infty}\frac{\arctan\left(\frac{1}{1+x+x^2}\right)}{\sin\left(\frac{1}{2x(x+1)}\right)}
$$
Now, use corollaries from the first remarkable limit:
$$
-\frac{1}{2}\lim_{x \to \infty}\frac{\arctan\left(\frac{1}{1+x+x^2}\right)}{\frac{1}{1+x+x^2}}\frac{\frac{1}{2x(x+1)}}{\sin\left(\frac{1}{2x(x+1)}\right)}\frac{\frac{1}{1+x+x^2}}{\frac{1}{2x(x+1)}}\to
-\frac{1}{2}\lim_{x \to \infty}\frac{\frac{1}{1+x+x^2}}{\frac{1}{2x(x+1)}}\to\\
-\frac{2}{2}\lim_{x \to \infty}\frac{x+x^2}{1+x+x^2}=-1
$$
Which implies:
$$
\lim_{x \to \infty} \frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}=-1
$$
A: Note that
$$\arctan(x+1)- \arctan(x) = \int_x^{x+1} \frac{dt}{1+t^2}.$$
A lower bound for this integral is $1/(1+(1+x)^2)\cdot 1,$ an upper bound is $1/(1+x^2)\cdot 1.$ We also have
$$\sin(1/x) - \sin (1/(x+1)) = \int_{1/(x+1)}^{1/x} \cos t\, dt.$$
Here a lower bound for the integral is $\cos (1/x)\cdot (1/x- 1/(x+1)),$ an upper bound is $\cos (1/(x+1))\cdot (1/x- 1/(x+1)).$ Noting $1/x- 1/(x+1) = 1/(x(1+x)),$ we can put all this together to see
$$\frac{1/(1+(x+1)^2)}{\cos (1/(x+1))(1/x(1+x))} \le \frac{\arctan(x+1)- \arctan(x)}{\sin(1/x) - \sin (1/(x+1))} \le \frac{1/(1+x^2)}{\cos (1/x)(1/x(1+x))}.$$
Both bounding functions $\to 1,$ hence so does the middle expression. Since I switched the sign of the denominator at the beginning (because it was easier for me to think about), the desired limit is $-1.$
A: 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 $x \to \infty$,

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

