Limit with inverse function 
If $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(x)=27x^3+(\cos 3+\sin 3)x$. Then $\displaystyle \lim_{x\rightarrow \infty}\frac{f^{-1}(1000 x)-f^{-1}(x)}{x^{\frac{1}{3}}}$ is ( where $f^{-1}(x)$ is inverse of $f(x)$)

Function $f(x)$ is strictly increasing function for all real $x$ 
So its inverse $f^{-1}(x)$ is also strictly increasing for all real $x$
Please have a look on that problem. I did not get any clue about that problem
Thanks
 A: Let $k=\cos 3+\sin 3$ so that $f(x) =27x^3+kx$. Further let $a, b$ be such that $f(a) =x, f(b) =1000x$ then $a, b$ are functions of $x$ which tend to $\infty $ with $x$. And we have to find the limit of $(b-a) /x^{1/3}$ as $x\to\infty $.
We have $$x=27a^3+ka,1000x=27b^3+kb\tag{1}$$ so that $$1000=\frac{b^3}{a^3}\cdot\frac{27+k/b^2}{27+k/a^2}$$ Taking limits as $x\to \infty$ so that $a, b$ also tend to  $\infty $ we see that $b/a\to 10$. And from $(1)$ we can see that $x/a^3\to 27$ or $x^{1/3}/a\to 3$. Now we have $$\frac{b-a} {x^{1/3}}= \frac{(b/a) - 1}{x^{1/3}/a}$$ and this tends to $$\frac{10-1}{3}=3$$

One should observe that the value of constant $k=\cos 3+\sin 3$ is immaterial here and is supposedly given to intimidate the reader. And if one somehow knows that the problem is independent of the value of $k$ a quick solution can be obtained by using $k=0$. Also it should be obvious via derivatives that the function $f$ is strictly increasing in the neighbourhood of $\infty$ so the problem makes sense. 
A: To check whether $f(x)=\alpha x^3+\beta x$ is strictly increasing, you would show that its derivative, $3\alpha x^2+\beta>0$, is positive for all $x$. Since $\alpha>0$, the derivative is a concave up quadratic with a minimum at $x=0$, as it is simply the quadratic $x^2$ scaled vertically by $3\alpha$ and translated vertically by $\beta$. Thus, in your case, you need only show that $3\cdot 27\cdot 0^2+(\cos 3 + \sin 3)>0\Rightarrow \sqrt{2}\sin (3+\frac{\pi}{4})>0$ by the double angle formula for $\sin$. But this is not the case, as $\sqrt{2}\sin(x+\frac\pi4)$ is negative in every interval $\left(\frac{(3+8k)\pi}{4},\frac{(7+8k)\pi}{4}\right)$ for $k\in\mathbb{Z}$ by considering the zeros of $\sin$ and the phase of the function. Thus, $\cos 3 + \sin 3$ is negative, as $\frac{3\pi}4<\frac{3\cdot 3.2}{4}<3<\frac{7\cdot3.1}{4}<\frac{7\pi}4$ so your $f$ is actually decreasing at $x=0$; not strictly increasing.
