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

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 $-1.$
