What's wrong with this argument? (Limits) In order to compute $\displaystyle \lim_{x \to \infty} \sqrt{9x^2 + x} - 3x$ we can multiply by the conjugate and eventually arrive at a limit value $1/6$. 
But what about the line of reasoning below, what is wrong with the argument and why? I can't think of a simple explanation, I had one involving the limit definition but I believe there should be a less complicated one. 
Here's the argument:
Clearly for large $x$ we can say $\sqrt{9x^2 + x} \approx \sqrt{9x^2} = 3x$. Hence $$ \lim_{x \to \infty} \sqrt{9x^2 + x} - 3x = \lim_{x \to \infty} 3x - 3x = 0 \ . $$ So the limit ought to be zero, easy!
What goes wrong and why?

The symbol $\approx $ is often misused.  It can mean numerical approximation or it can mean asymptotically equal.
For example it is not true that (n+1) 2  $\approx $  n 2  in both senses. This you can see from the difference 
${(n + 1)^2} - {n^2} = 1 + 2n$ .
You can observe that the difference is not small. Indeed it tends to infinity as n tends to infinity. 
In $\mathop {\lim }\limits_{x \to \infty } \sqrt {9{x^2} + x}  - 3x$ you are taking the limit as x tends to infinity. You are looking for the difference of 
$\sqrt {9{x^2} + x} $ and $ 3x $ as x gets larger and larger.  The limit exists if this difference gets closer and closer to a certain value.
In example in variable $n$ above, the difference tends to infinity and so no limit.
If you now look at the difference 
$\sqrt {9{x^2} + x}  - 3x = \frac{{\left( {\sqrt {9{x^2} + x}  - 3x} \right)}}{{\sqrt {9{x^2} + x}  + 3x}}\left( {\sqrt {9{x^2} + x}  + 3x} \right) = \frac{x}{{\sqrt {9{x^2} + x}  + 3x}}$,  you see that 
$\sqrt {9{x^2} + x}  = 3x + \frac{x}{{\sqrt {9{x^2} + x}  + 3x}}$ 
and 
$\frac{x}{{\sqrt {9{x^2} + x}  + 3x}} \ge \frac{x}{{\sqrt {16{x^2}}  + 3x}} = \frac{1}{7}$ no matter how large x gets and $1/7$ is definitely not a small value. So you cannot say 
$\sqrt {9{x^2} + x}  \approx 3x$ .
It is precisely this difference that gives you the limiting value. This question is to let you know it is erroneous to think of estimating large value as approximation. It is possible two functions need not have the same value as x gets larger and larger and it is possible that the two functions for large values of x are always a fixed quantity apart, or gets  further and further  apart or their difference may oscillate about a fixed quantity.  For example
x 2 and (x+1)2,  x2 and x2+sin(x), x2 and x2+6. 
More precislely for this question, you want to show that no matter how small $\epsilon $ is you can find number $K$ so that for all $x > K$ , the difference
$\left| {\sqrt {9{x^2} + x}  - 3x - \frac{1}{6}} \right| = \left| {\frac{x}{{\sqrt {9{x^2} + x}  + 3x}} - \frac{1}{6}} \right| < \varepsilon $   
This is equivalent to 
$\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {9{x^2} + x}  + 3x}} = \frac{1}{6}$ .
This I think you can prove. 
Now you can say $\sqrt {9{x^2} + x}  \approx 3x + \frac{1}{6}$ 
or more precisely $\sqrt {9{x^2} + x} $ is asymptotically equal to $3x + \frac{1}{6}$.
