\(\begin{array}{l}
\quad f(x) = \dfrac{x}{\sqrt{x^2 - 1}}\\
a)\quad \lim\limits_{x\to +\infty}f(x)\\
= \lim\limits_{x\to +\infty}\dfrac{x}{\sqrt{x^2 - 1}}\\
= \lim\limits_{x\to +\infty}\dfrac{1}{\sqrt{1 - \dfrac{1}{x^2}}}\\
= \dfrac{1}{\sqrt{1 - 0}}\\
= 1\\
b)\quad [f(x)]'\\
= \left[\dfrac{x}{\sqrt{x^2 - 1}}\right]'\\
= \dfrac{\sqrt{x^2 - 1} - x\cdot \dfrac{x}{\sqrt{x^2 - 1}}}{x^2 - 1}\\
= \dfrac{x^2 - 1 - x^2}{(x^2 - 1)\sqrt{x^2 - 1}}\\
= - \dfrac{1}{\sqrt{(x^2 - 1)^3}}\\
c)\quad \displaystyle\int\limits_\sqrt2^bf(x)dx\quad \left(b > \sqrt2\right)\\
= \displaystyle\int\limits_\sqrt2^b\dfrac{x}{\sqrt{x^2 - 1}}dx\\
= \displaystyle\int\limits_\sqrt2^b\dfrac{d(x^2 - 1)}{2\sqrt{x^2 - 1}}\\
= \sqrt{x^2- 1}\Bigg|_\sqrt2^b\\
= \sqrt{b^2 - 1} - 1
\end{array}\)
