Question

\[\lim_{x \rightarrow \infty} \sqrt{9x^2+x}-3x\]

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Actually nevermind, it's +x.

Answer 5

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

Answer 6

Oh ok.

Answer 7

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