# Ask Uncle Colin: A Short, Sweet Limit

Dear Uncle Colin,

What is $\lim_{x \to \infty} \left\{ \sqrt{x^2 + 3x} - x\right\}$?

- Raging Over Obnoxious Terseness

Hi, ROOT, and thanks for your very brief question.

My approach would be to split up the square root and use either a binomial expansion or completing the square, as follows:

$\sqrt{x^2 + 3x} = x\sqrt{1 + \frac{3}{x}}$.

The binomial expansion gives $\sqrt{1+\frac{3}{x}} \approx 1 + \frac{3}{2x}$, so $\sqrt{x^2 + 3x}-x \approx x(1 + \frac{3}{2x}) - x = \frac{3}{2}$.

If you don’t like the binomial expansion, you can also set up the square root as $\sqrt{\left(x+\frac{3}{2}\right)^2 - \frac{9}{4}}$, which becomes $\left(x + \frac{3}{2}\right)\sqrt{1 - \frac{9}{4\left(x+\frac{3}{2}\right)^2}}$. As $x \to \infty$, the square root goes to 1 and you’re left with $\left(x+\frac{3}{2}\right)-x$, or $\frac{3}{2}$.

Hope that helps!

-- Uncle Colin