## hpfan101 one year ago $\lim_{x \rightarrow \infty} \sqrt{9x^2+x}-3x$

1. anonymous

gimmick is to multiply by $\frac{\sqrt{9x^2+x}+3x}{\sqrt{9x^2+x}+3x}$

2. hpfan101

When I multiplied by the conjugate I got: $\frac{ 9x^2+x-9x }{ \sqrt{9x^2+x}+3x }$

3. hpfan101

Oh the denominator is supposed to be -x not +x

4. hpfan101

Actually nevermind, it's +x.

5. anonymous

$$-9x^{\color{red}2}$$ in the numerator's second term

6. hpfan101

Oh ok.

7. hpfan101

Alright, I now know how to solve the question from there. Thank you guys!