is this line of thinking valid for quick solving? I came across a seemingly simple limit when looking for an oblique assymptote of a function:
$(x^2)/(\sqrt{x^2-1})-x$

I need to know what happens here as $x$ approached infinity. The usual method of dividing by the highest degree polynomial doesn't work here, and its become time-consuming relative to the larger problem that this is associated with. 
So I just decided to look at things this way:
when $x$ approaches infinity $\sqrt{x^2-1}$ and $\sqrt{x^2}$ are pretty much the same thing, since a measly $-1$ can't have much effect in infinity. 
And then I simplified the limit to:
$(x^2/x) - x = 0$ 
Is this totally wrong? Is it by chance that I got the correct limit?
I'm sure there are great ways of proving this limit rigorously but my question is if I come across such a limit and need to quickly move on can I make these assumptions?
Thanks
 A: It is indeed wrong. You can factor your expression like that (considering $x \geq 0$):
$$ {x^2 \over \sqrt{x^2 - 1}} - x = x \left({1 \over \sqrt{1 - {1 \over x^2}}} - 1 \right) $$
The notion that $\sqrt{x^2-1}$ is "about equal to" $\sqrt{x^2}$ is captured by the notion of asymptotic equivalence (more info is available at Wikipedia). The problem with that is that equivalence doesn't behave well with addition, and even if $f(x) \sim g(x)$ then you don't necessarily have $f(x) + h(x) \sim g(x) +h(x)$. This is what happen here.
But you can use a polynomial approximation of your function to compute the actual limit. We know that $(1+x)^\alpha = 1 + \alpha x + o(x)$ when $x \rightarrow 0$. Using that you get $\left( 1 - {1 \over x^2} \right)^{-1/2} = 1 + {1 \over 2 x^2} + o({1 \over x^2})$ as $x \rightarrow \infty$, so your expression is equal to ${1 \over 2x} + o({1 \over x})$. This does converge to zero as $x\rightarrow\infty$, but this required a better argument than the one you gave.
You could easily construct a similar example which doesn't converge to zero: take $f(x) = {x^3 \over \sqrt{x^2-1}} - x^2$, then by a similar argument $f(x) = 
{1 \over 2} + o(1)$ which converges to ${1 \over 2}$ as $x \rightarrow \infty$, even though is you just used $\sqrt{x^2-1} \simeq x$ you'd get zero. Multiplying again by $x$ you get something that diverges to $\infty$.
A: For the impatient: You can skip almost all of this post, and go to the last $4$ lines. 
The informal approach described by the OP almost works, but needs a little tweaking.  We start informally, but end up with a quick formal proof.  I want to tell the story in a roundabout way, in order to begin to convey the thinking.
Imagine $x$ large positive.  It is clear that $\sqrt{x^2-1}$ is smaller, but not much smaller, than $x$, so $\frac{x^2}{\sqrt{x^2-1}}$ is bigger than $x$, and therefore 
$$\frac{x^2}{\sqrt{x^2-1}}-x >0.$$
This is not enough to tell us the limit. We need to answer, at least roughly, the question: $\sqrt{x^2-1}$ is how much smaller than $x$?.
Maybe smaller by $u$, where $u$ is small?  How small?  We have 
$$\sqrt{x^2-1}=x-u,$$
and therefore 
$$x^2-1=x^2-2ux+u^2.$$
or more simply $2ux-u^2=1$.
Since $u$ is small, $2ux$ is very much bigger than $u^2$.  So $2ux\approx 1$, and therefore $u\approx \frac{1}{2x}$.  This is an excellent approximation. (By the way, it is the tangent line approximation.)
So, informally, 
$$\frac{x^2}{\sqrt{x^2-1}}-x\approx \frac{x^2}{x-\frac{1}{2x}}-x=\frac{1}{2}\frac{1}{x-\frac{1}{2x}}.$$
Great, looks like we are finished.  It is obvious that the right-hand side is tiny when $x$ is huge.  We even have good control over how tiny. 
Unfortunately, for very formal purposes, there is a little problem. For note that if $u=1/(2x)$ then $(x-u)^2$, though a wonderful approximation to $x^2-1$, is unfortunately a little bit bigger than $x^2-1$.  
Let's fix that, crudely, because crude is good enough. Instead of $1/(2x)$, we will use $1/x$.
$$\left(x-  \frac{1}{x}\right)^2 =x^2-2+\frac{1}{x^2}.$$  For $x$ large, $x^2-2+\frac{1}{x^2}$ is clearly less than $x^2-1$.  So now we know that 
$\frac{x^2}{\sqrt{1-x^2}}<\frac{x^2}{x-\frac{1}{x}}$. That upper bound will do the job. 
Proof of the limit result: We now write up a solution, carefully erasing all traces of the thinking that went into it.
It is easy to verify that if $x>1$, then $(x-1/x)^2 <x^2-1$.  It follows that 
$$0 <\frac{x^2}{\sqrt{x^2-1}}-x<\frac{x^2}{x-\frac{1}{x}}-x=\frac{1}{x-\frac{1}{x}}.$$
By Squeezing, it follows that
$$\lim_{x\to\infty}\left(\frac{x^2}{\sqrt{x^2-1}}-x\right)=0.$$
A: $$\displaystyle \lim_{x \to \infty} \frac{x^2-x\sqrt{x^2-1}}{\sqrt{x^2-1}} =\displaystyle \lim_{x \to \infty}  \frac{x(x-\sqrt{x^2-1})}{\sqrt{x^2-1}}\cdot \frac{(x+\sqrt{x^2-1})}{(x+\sqrt{x^2-1})}=$$
$$= \lim_{x \to \infty} \frac{x}{\sqrt{x^2-1}\cdot (x+\sqrt{x^2-1})} = \lim_{x \to \infty} \frac{1}{\sqrt{1-\frac{1}{x^2}}\cdot (x+\sqrt{x^2-1})}=\frac{1}{\infty}=0$$
