Limit of a square root with $x^2+x$

Question

I saw the following problem: $$\lim_{x\to \infty} \sqrt{9x^2+x}-3x$$ My first thought was to say that the $x$ term is overpowered when $x$ becomes large enough, so the square root becomes just $\sqrt{9x^2} = 3x$, and the value of the limit is zero.

However, the solution given is $1\over6$, and starts with: $$\lim_{x\to \infty} \sqrt{9x^2+x}-3x=\lim_{x\to \infty} {(\sqrt{9x^2+x}-3x)(\sqrt{9x^2+x}+3x)\over \sqrt{9x^2+x}+3x}$$

I'm assuming the continuation is:

$$=\lim_{x\to \infty} {x\over \sqrt{9x^2+x}+3x} = \lim_{x\to \infty} {x\over \sqrt{9x^2}+3x} = \lim_{x\to \infty} {x \over 6x} = {1 \over 6}$$

The question is, why is it okay to ignore the $x$ term in $\sqrt{9x^2+x}+3x$ but not in $\sqrt{9x^2+x}-3x$?

Answer 1

$$\frac{x}{\sqrt{9x^2 + x} + 3x} = \frac{1}{\sqrt{9+\frac{1}{x}}+3}$$

Answer 2

You should note that $9x^2+x = (3x+\frac 16)^2-\frac 1{36}$ so that $\sqrt {9x^2+x}\approx 3x+\frac 16$ for large $x$

When you subtract $3x$ in the numerator, the dominant term in $x$ vanishes and the highest order term is $\frac 16$.

In the denominator, you add $3x$ and the highest order term is $6x$.

The use of the conjugate in this way - where the highest order term in the numerator will cancel - is quite common, and is useful in other circumstances too. Here it avoids having to complete the square or justify the approximation.

Answer 3

Because the original limit is an indeterminate form, where that $x$ moves you a bit further from $0$. At any rate, formally you're not allowed to simply drop a term. Initially, as you can see, the fundamental law of fractions was used to rationalize the numerator. In the latter case, factorizing the denominator leads to the answer: $$\lim_{x\to \infty} {x\over \sqrt{9x^2+x}+3x}=\lim_{x\to \infty}\frac{x}{x\sqrt{9+1/x}+3x}=\lim_{x\to \infty}\frac{x}{x(\sqrt{9+1/x}+3)}=\\ \lim_{x\to \infty}\frac{1}{\sqrt{9+1/x}+3}=\frac{1}{6}. $$