I have a question about the limit to infinity of a rational expression with both a radical and non-radical in the denominator. It may be possible that I am not following the correct steps for this problem, I was hoping someone could explain.
I start with the following expression:
$$\lim_{x\to \infty}\sqrt{9x^2+x}-3x$$
I multiply it by its conjugate to get:
$$\frac{x}{\sqrt{9x^2+x}+3x}$$
And this is where I'm stuck. Normally at this point I would identify $9x^2$ in the denominator as the highest term, multiply the numerator by $\frac{1}{x^2}$ and multiply the denominator by $\frac{1}{\sqrt{x^4}}$. However there is a pesky $+3x$ in the denominator and I don't know how to deal with it.
Best Answer
Hint: We have for positive $x$ that $\sqrt{9x^2+x}=3x\sqrt{1+\frac{x}{9x^2}}$. Now divide top and bottom by $x$. The pesky $3x$ behaves like a lamb.