I am trying to find the limit and I know that I have to do something to get a denominator but I can't make it work. There is a specific thing I have to do to this problem to make it work but I am not sure what that is. I know what I am doing is not wrong mathematically, it just doesn't help me get the answer.
$$\lim_{x \to \infty} (\sqrt{9x^2+x} – 3x)$$
Not sure where to go with this, multiply it all by the conjugate? I tried that but I know that won't help because I have to rationalize the denominator. I know I can't multiply by anything really until I get rid of the sqrt and I don't know how.
Best Answer
$$\lim_{x\to\infty}(\sqrt{9x^2+x}-3x)$$ $$=\lim_{x\to\infty}\frac{x}{\sqrt{9x^2+x}+3x}.$$ Divide numerator and denominator by $x$, $$=\lim_{x\to\infty}\frac{1}{\frac{1}{x}\sqrt{9x^2+x}+3}$$ note $\frac{1}{x}\sqrt{9x^2+x}=\sqrt{\frac{9x^2+x}{x^2}}=\sqrt{9+\frac{1}{x}}$ so this reduces to $$=\lim_{x\to\infty}\frac{1}{\sqrt{9+\frac{1}{x}}+3}$$ $$=\frac{1}{\sqrt{9}+3}=\frac{1}{6}.$$ And just FYI this ground was (more or less) covered in this question.