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Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.
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