Finding Horizontal/Oblique Asymptote of $y=\frac{\sqrt{x}+1}{\sqrt{x}-1}$ Is it correct to simply subsitute $\sqrt{x}$ with $x$ when finding horizontal or oblique asymptotes? The method works but I am not sure if it is formally sound enough to pass muster in an examination.
How about a function of other fractional powers of $x$. If I had a function where $x$ always appeared as a seventh root: $\sqrt[7]{x}$, can I substitute $\sqrt[7]{x}$ with $x$ to find asymptotes?
If so, in general, can I substitute $g(x)$ with $x$ to find the asymptotes of $f(g(x))$ if $\lim_{x \to \infty}g(x) = \infty$?
 A: Hint
You can write $$y=\frac{\sqrt[n]{x}+1}{\sqrt[n]{x}-1}=1+\frac{2}{\sqrt[n]{x}-1}$$ So, if $x$ goes to infinity, you have $$y \simeq 1+\frac{2}{\sqrt[n]{x}}$$
A: I would try to express as a mixed fraction instead.$
$$\frac{\sqrt{x}+1}{\sqrt{x}-1} = \frac{\sqrt{x}+(-1+1)+1}{\sqrt{x}-1}=\frac{(\sqrt{x}-1)+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1} $$
A: The answer is a firm No. In the following explanation, please ignore the issue of complex roots because it will distract from the bigger problem with the method proposed.
Consider $y=\frac{x+1}{\sqrt{x}-1}$. Since it can be expressed entirely in terms of $\sqrt{x}$, the method proposed in the question is to simply substitute it with $x$ and proceed to find the asymptotes.
The proposed substitution yields $y=\frac{x^2+1}{x-1}$ which can be simplified to $y=x+1+\frac{2}{x-1}$. As $x$ goes to $\infty$, this becomes $y=x+1$. It might be tempting to conclude that $y=x+1$ is the oblique asymptote as suggested by the method proposed in the question.
However, this is wrong because the correct end result of the limit evaluation should be to substitute $\sqrt{x}$ back in place of $x$ yielding $y=\sqrt{x}+1$ which is not an oblique asymptote. (If this is confusing, use a different variable in the previous step.)
Thus, the method proposed by the question is wrong in general. It may suggest oblique asymptote of the general form: $y=m.g(x)+c$ which will not be linear in general and therefore not an asymptote.
The only situation in which the method works is when the given function has only a horizontal asymptote, i. e. where $m=0$, or when $g(x)$ is linear but a linear $g(x) is not very interesting anyway.
