Let's denote $\,I(x)=\displaystyle \int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt$
Integrating by part
$$I(x)=t\tan^{-1}\left( t^x-t^{-x} \right)\Big|_0^1-x\int_{0}^{1}\frac{\left( t^x+t^{-x} \right)}{1+\left( t^x-t^{-x} \right)^2}dt$$
Making the substitution $t=e^{-s}$
$$=-x\int_0^\infty\frac{\left( e^{sx}+e^{-sx}\right)}{1+\left( e^{sx}-e^{-sx}\right)^2}e^{-s}ds\overset{t=sx}{=}-2\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}e^{-\frac{t}{x}}dt$$
With the accuracy up to exponentially small corrections at $x\gg1$, we can decompose the exponent: $\displaystyle e^{-\frac{t}{x}}=1-\frac{t}{x}+\frac{t^2}{2x^2}+O\Big(\frac{1}{x^3}\Big)$
$$I(x)=-2\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}dt+\frac{2}{x}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}t\,dt+O\Big(\frac{1}{x^2}\Big)$$
$$=-\frac{\pi}{2}+\frac{2}{x}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t\,dt-\frac{1}{x^2}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t^2\,dt+O\Big(\frac{1}{x^3}\Big)$$
and, therefore, the desired limit
$$\lim_{x \to \infty} x\left( 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt \right)=\frac{4}{\pi}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t\,dt=0.77749...$$
(it can also be presented in the form $\displaystyle -\,\frac{2}{\pi}\int_0^1\frac{1+t^2}{1-t^2+t^4}\ln t\,dt\,$).
The numeric evaluation with WolframAlpha confirms the answer.
$\bf{Update}$
As @170231 mentioned, the last integral has a closed form:
$$ J=\int_0^1\frac{1+t^2}{1-t^2+t^4}\ln t\,dt=-\,\frac{4}{3}G$$
and the desired limit is
$$\boxed{\,\,\lim_{x \to \infty} x\left( 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt \right)=\frac{8G}{3\pi}=0.77749...\,\,}$$
Indeed,
$$J=\int_0^1\frac{(1+t^2)^2}{1+t^6}\ln t\,dt=$$
$$\int_0^1(1-t^6+t^{12}-...)\ln t\,dt+2\int_0^1(t^2-t^8+t^{14}-...)\ln t\,dt$$
$$+\int_0^1(t^4-t^{10}+t^{14}-...)\ln t\,dt$$
$$=-1+\frac{1}{7^2}-\frac{1}{13^2}+..+2\Big(-\frac{1}{3^2}+\frac{1}{9^2}-\frac{1}{15^2}-..\Big)-\frac{1}{5^2}+\frac{1}{11^2}-\frac{1}{17^2}+...$$
Denoting the Catalan' constant $\displaystyle G=1-\frac{1}{3^2}+\frac{1}{5^2}-...$
$$J=-1+\frac{1}{3^3}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{9^2}+ ... -\frac{2}{9}G-\frac{1}{3^2}+\frac{1}{9^2}-\frac{1}{15^2}+...$$
$$=-G-\frac{2}{9}G-\frac{1}{9}G=-\,\frac{4}{3}G$$
It is interesting to note that the second term of the asymptotics $I(x)$ can also be found in a closed form:
$$I(x)=I_0+\frac{1}{x}I_1+\frac{1}{x^2}I_2\,+...$$
where
$$I_2=-\int_0^\infty\frac{t^2\cosh t}{1+4\sinh^2t}\,dt=-\frac{1}{2}\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}\int_{-\infty}^\infty\frac{e^{-\beta t}\cosh t}{1+4\sinh^2t}dt=-\frac{1}{2}\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}J(\beta)$$
The last integral can be evaluated by means of integration along a rectangular contour in the complex plane:
$$J(\beta)=\frac{\pi}{2}\,\frac{\cos\frac{\beta\pi}{3}}{\cos\frac{\beta\pi}{2}};\quad\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}J(\beta)=\frac{5\pi^3}{72}$$
and
$$I(x)=-\frac{\pi}{2}+\frac{4G}{3}\,\frac{1}{x}-\frac{5\pi^3}{144}\,\frac{1}{x^2}+O\Big(\frac{1}{x^3}\Big)$$
Best Answer
You have $1^{\infty}$ form. so use this :-
$$\lim_{x\to a}f(x)^{g(x)} = \exp(\lim_{x\to a} (f(x)-1)g(x))$$ when $\lim_{x\to a}f(x)^{g(x)}$ is $1^{\infty}$ form.
Here is a proof for the above
So you have :-
$$\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{2}\sin(x)})=\\\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}}\frac{x}{\sin(x)})=\\\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}})$$.
Now use L'Hospital:-
$$\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}})=\exp(\lim_{x\to 0}\frac{1-\cos(x)}{3x^{2}})=\\\exp(\lim_{x\to 0}\frac{\sin(x)}{6x})=\exp(\frac{1}{6})$$.
Giving you the final answer as $$e^{\frac{1}{6}}$$