General method to find the oblique asymptotes to a curve All the books I've looked at say that if $y=mx+c$ is an asymptote, then the limit as $x$ tends to infinity is $y/x =m$. 
So does that also mean that, as $x$ tends to infinity, $y$ also tends to infinity? (I think that it does because that is what makes an asymptote and this fact is used later in the question. If this is correct why don't we then write $m=\lim_{y\to\infty} y/x$?). Edit please clarify this point because the comment says it's incorrect.
I  couldn't write the question here so it's in the pictures below. 


Questions:


*

*Using my method, how can I find the $c$ in $y=mx+c$ after finding $m$?

*Why has the author first substituted $y=mx+c$ in the equation of the curve?
Edit: is it because the asymptote will touch the curve at infinity

*Why does the author say to equate the coefficients of the first $2$ highest degree terms of $x$ (i.e $x^n$ and $x^{n-1}$)?


Is there also some other way to solve this problem?
Edit : The biggest problem is why Meadra substituted x=0 after substituting y=2x + c in the equation of the curve. She should have substituted x=0 in y=2x + c because the asymptote would cut the curve or intersect at x tending  to infinity so  at x is equal to zero curve will not be similar to the asymptote
 A: Sorry for the debate down,I figured out the result this time 
Consider the intersection of the surface $y^3 -6xy+11x^2y-6x^3+x+y$ with the $z$ plane we have the curve defined by  $y^3 -6xy+11x^2y-6x^3+x+y = 0(1)$.The function is defined implicitly so consider an asymptote to the curve it has the form $y=mx+c$ 
Dividing $(1)$ by $x^3$ and subsituting $y=mx+c$ We have
$$\displaystyle \frac{(mx+c)^3}{x^3} -6\frac{mx+c}{x^2}+11\frac{mx+c}{x}-6+\frac{mx+c}{x^3} = 0$$
Letting $x \to \infty$ we get. $12m=6$ Thus $m=2$
Thus the asymptotes are of the form $y = 2x+c$.Subsitute in $(1)$
$$3x^3 +12x^2c +6xc^2+c^3+12x^2+6xc+22x^3+11x^2-12x^4-6x^3c+x+2x+c=0$$
Now when $x=0$(I do this because since I subsituted the eqn of the asymptote in (1) the resulting equation varies the same way as the asymptote and to find the intercept we just let $x=0$) we get $y=c$ thus we have $c^3+c=0$ thus $c=0$
As a result the asymptote to the curve is $y=2x$
A: now solve for $m^3 - 6m^2 + 11m - 6 = 0$, you get three $m_1 = 1$, $m_2 = 2$, and $m_3 = 3$. Then, you have $y = x$, $y = 2x$, and $y = 3x$. Therefore, you have three asymptotes, which means that the oblique asymptote does not exist.
