# Thread: the limit as x approaches infinity

1. ## the limit as x approaches infinity

I know the answer to this problem, but I do not know how to get to it.

lim (sqrt(9x^2+x)-3x) = (1/6)
x-> infinity

2. Multiply top and bottom by its conjugate: $\sqrt{9x^2 + x} {\color{red}+} 3x$:

$\lim_{x \to \infty} \left(\sqrt{9x^2 + x} - 3x\right) \cdot \frac{{\color{red}\sqrt{9x^2 + x} + 3x} }{{\color{red}\sqrt{9x^2 + x} + 3x}}$

The numerator is an application of the difference of squares formula: $(a-b)(a+b) = a^2 - b^2$

So: $= \lim_{x \to \infty} \frac{9x^2 + x \ - \ 9x^2}{\sqrt{9x^2 + x} + 3x}$

etc. etc.

3. $\frac{\sqrt{9x^2 + x} - 3x}{1} \cdot \frac{\sqrt{9x^2 + x} + 3x}{\sqrt{9x^2 + x} + 3x}$

$\frac{(9x^2 + x) - 9x^2}{\sqrt{9x^2 + x} + 3x}
$

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

divide every term by x ...

$\frac{1}{\sqrt{9 + \frac{1}{x}} + 3}$

now let $x \to \infty$

4. Hello,

Another approach is to use asymptotic equivalences (but I don't know your level, so maybe you can't use it...)

$\sqrt{9x^2+x}-3x=3x \left(\sqrt{1+\frac{1}{9x}}-1\right)$

Since $x \to \infty$, we have $\frac{1}{9x} \to 0$

$\sqrt{1+\frac{1}{9x}}=\left(1+\frac{1}{9x}\right)^ {1/2} \sim_{\substack{x \uparrow \infty}} 1+\frac 12 \times \frac{1}{9x}+ \mathcal{O}\left(\frac{1}{x^2}\right)$

So $\lim_{x \to \infty} \sqrt{9x^2+x}-3x=\lim_{x \to \infty} 3x\left(1+\frac 12 \times \frac{1}{9x}+ \mathcal{O}\left(\frac{1}{x^2}\right)-1\right)$

$=\lim_{x \to \infty} 3x \left(\frac 12 \times \frac{1}{9x}+\mathcal{O} \left(\frac{1}{x^2}\right)\right)=\lim_{x \to \infty} \frac 16+\mathcal{O} \left(\frac 1x\right)=\boxed{\frac 16}$