find oblique asymptotes : $f(x) =\sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}$ let : $$f(x) =\sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}$$
shuch that : $$m-n=q$$
then : fine oblique asymptotes :
i know that : $y=mx+h$ is oblique asymptotes  such that :
$$m=\lim_{x\to \infty} \frac{f(x)}{x}$$
and :
$$h=\lim_{x\to \infty}(f(x)-mx)$$
But I can not answer 
please help .

For the slope, you can do the same calculation you mentioned:
$$\begin{align}
\displaystyle\lim_{x\to\infty}\frac{f(x)}{x} &= \displaystyle\lim_{x\to\infty} \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1} + \cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}\cdot\frac{1}{x}\\
&= \displaystyle\lim_{x\to\infty} \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1} + \cdots+b_0}}\cdot\sqrt[q]{\frac{1}{x^q}}\\
&= \displaystyle\lim_{x\to\infty} \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^m+b_{n-1}x^{m-1} + \cdots+b_0x^q}}\\
&= \sqrt[q]{\displaystyle\lim_{x\to\infty}\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^m+b_{n-1}x^{m-1} + \cdots+b_0x^q}}\\
&= \sqrt[q]{\frac{a_m}{b_n}}
\end{align}$$
Similarly:
$$h = \displaystyle\lim_{x\to\infty}f(x) - (\text{slope})x = \displaystyle\lim_{x\to\infty} \left( \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1} + \cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}} - \sqrt[q]{\frac{a_m}{b_n}}x \right)$$
This limit is trickier. Do you know any methods for evaluating it?

Let's try this: 
$$\begin{align}
\sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1} + \cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}} &= \left(\sqrt[q]{\frac{a_m}{b_n}}\right)x \cdot \sqrt[q]{\frac{x^n + \frac{a_{m-1}}{a_m}x^{n-1} + \cdots + \frac{a_0}{a_m}x^{-q}}{x^n + \frac{b_{n-1}}{b_n}x^{n-1} + \cdots + \frac{b_0}{b_n}}}\\
&= \left(\sqrt[q]{\frac{a_m}{b_n}}\right)x \cdot \left(1 + \left(\frac{a_{m-1}}{a_m} - \frac{b_{n-1}}{b_n}\right)x^{-1} + \cdots\right)^{1/q}
\end{align}$$
where everything in the dots at the end involves powers $x^{-2}$ or smaller. Now, close to $x=1$, we know that $(1+h)^{1/q}$ is well-approximated by $1+\frac{h}{q}$. Thus, we replace the complicated factor on the right with an approximation:
$$\left(\sqrt[q]{\frac{a_m}{b_n}}\right)x \cdot \left(1 + \frac{1}{q}\left(\frac{a_{m-1}}{a_m} - \frac{b_{n-1}}{b_n}\right)x^{-1} + \cdots\right)$$
Multiplying this out, we get that
$$f(x) = \left(\sqrt[q]{\frac{a_m}{b_n}}\right)x + \sqrt[q]{\frac{a_m}{b_n}}\cdot\frac{1}{q}\left(\frac{a_{m-1}}{a_m} - \frac{b_{n-1}}{b_n}\right) + O(x^{-1})$$
Based on this, and encouraged by numeric experiments, I'm pretty sure that we can say 
$$h=\sqrt[q]{\frac{a_m}{b_n}}\cdot\frac{1}{q}\left(\frac{a_{m-1}}{a_m} - \frac{b_{n-1}}{b_n}\right).$$
I'd love to see a simpler way to get there. I got this idea from one of the answers to: How to find the oblique asymptote of root of a function?
