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?
 A: A more accurate way to perform an approximation would be to complete the square:
\begin{align*}
\sqrt{9x^2 + x} - 3x
&= \sqrt{9x^2 + x + \frac{1}{36} - \frac{1}{36}} - 3x \\
&= \sqrt{\left(3x + \frac{1}{6} \right)^2 - \frac{1}{36}} - 3x \\
&\approx \sqrt{\left(3x + \frac{1}{6} \right)^2} - 3x \\
&= \frac{1}{6}
\end{align*}
A: The simple answer to your question is that the reasoning is not correct because, as you've proven through some algebraic manipulation, the limit is $\frac{1}{6}$ and limits are unique in this case. 
More generally though, your reasoning confuses two notions of closeness. While it is true that: 
$$ \lim_{x \to \infty} \frac{9x^2 + x}{9x^2} = 1$$
it is evidently not true that: 
$$ \lim_{x \to \infty} (9x^2 + x) = \lim_{x \to \infty} 9x^2 $$ since this is equivalent to: 
$$ \lim_{x \to \infty} x = 0 $$
which would be true if: 
$$ \lim_{x \to \infty} \sqrt{9x^2 + x} = \lim_{x \to \infty} \sqrt{9x^2} $$. 
A: The issue arises because of a lack of understanding of the symbol $\approx$ used here. It is very important in mathematics to work out things rigorously even if one is not using a formal notation. The meaning of the symbol $\approx$ as used here is that $$f(x) \approx g(x)\text{  when }x\to a\text{ if } \lim_{x \to a}\frac{f(x)}{g(x)} = 1\tag{1}$$ This does not guarantee that $$\lim_{x \to a}\{f(x) - g(x)\} = 0\tag{2}$$ However if $\lim_{x \to a}g(x)$ exists and is non-zero then the result $(2)$ follows from $(1)$ (this can be easily proved using rules of algebra of limits). In this specific case we have $x \to \infty$ and $f(x) = \sqrt{9x^{2} + x}, g(x) = 3x$ and clearly $\lim_{x \to \infty}g(x)$ does not exist. Hence we can't go from $(1)$ to $(2)$ and hence the expression $\sqrt{9x^{2} + x}$ can't be replaced by $3x$ while evaluating the limit of $\sqrt{9x^{2} + x} - 3x$ when $x \to \infty$.
However the reasoning applied by OP is very common among students and I believe that this is more of a pedagogic issue. Teaching calculus is really a difficult game and hence many instructors try to make things simplistic (or we say intuitive) even at the expense of rigor. Thus $a \approx b$ means that $a$ is nearly equal to $b$ (how near?) instead of the fact that $a/b$ is nearly equal to $1$. I have found that unless one is experienced in the art of calculus, intuition does not help much in calculus and it is better to stick to simple rules of limits rather than thinking in such vague terms.
A: Here, Limit deals with least upper bound... but u have chosen $\sqrt{9x^2+x}\approx 3x$. But we know that $3x\leq \sqrt{9x^2+x}$ for all values of $x$...So u can conclude only that $$\lim_{n\to \infty}\sqrt{9x^2+x}-3x\geq\lim_{n\to \infty}(3x-3x)= 0$$ but can't conclude that equal to $0$.
A: I would argue in the following direction:
When you write that $\sqrt{9x^2 +x} \sim \sqrt{9x^2}$ you are saying that 
$$\lim_{x \rightarrow \infty} \frac{\sqrt{9x^2 +x}}{\sqrt{9x^2}}=1$$
This equivalence means that, if you take $x$ large enough, you get 
$$ \sqrt{9x^2}(1-\epsilon)< \sqrt{9x^2 +x} < \sqrt{9x^2}(1+\epsilon)$$ 
Observe that you have an error of $\epsilon\sqrt{9x^2}$ that remains when you subtract $3x$. This shows you only that $-\epsilon\sqrt{9x^2}<\sqrt{9x^2 +x}-3x<\epsilon\sqrt{9x^2}$ what isn´t enough to "prove" that the limit is zero.
A: The theorem say: if $f(x)\leq g(x)$, then $\lim_{x\to a}f(x)\leq \lim_{x\to a}g(x)$, note that:
$$9x^2\leq9x^2+x\implies\sqrt{9x^2}\leq \sqrt{9x^2+x}\implies \sqrt{9x^2}-3x\leq \sqrt{9x^2+x}-3x$$
Take $f(x)=\sqrt{9x^2}-3x$, and $g(x)=\sqrt{9x^2+x}-3x$, and $a=\infty$.
A: 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 $3x + \frac{1}{6}$.
