# Limits

September 26th, 2009 by ben

Question from the tryout:

Find:

$\dfrac{\displaystyle\lim_{x \to 0} \dfrac{\sin \dfrac{x}{5}}{\sin 2x} \cdot \lim_{x \to 3^{-}} \lfloor x \rfloor}{\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} + \lim_{x \to\infty} \dfrac{3 - 4x^2}{2x^2 + 7}}$

This problem just looks scary, doesn’t it? Well, let’s split up the four limits.

First limit, the top left.

$\displaystyle\lim_{x \to 0} \dfrac{\sin \dfrac{x}{5}}{\sin 2x}$

l’Hôpital it! Bwahaha.

$\displaystyle\lim_{x \to 0} \dfrac{\dfrac{1}{5}\cos \dfrac{x}{5}}{2 \cos 2x} = \dfrac{\dfrac{1}{5}}{2} = \boxed{\dfrac{1}{10}}$

Now for the top right:

$\displaystyle\lim_{x \to 3^{-}} \lfloor x \rfloor$

The limit of the floor function approaching from the negative side is the lower number, so this equals $\boxed{2}$.

Bottom left:

$\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$

Indeterminate form (zero over zero), so once again we can use l’Hôpital’s rule on it.

$\displaystyle\lim_{x \to 3} \dfrac{2x}{1} = \boxed{6}$

Last limit is the bottom right one.

$\displaystyle\lim_{x \to\infty} \dfrac{3 - 4x^2}{2x^2 + 7}$

This one is $\frac{\infty}{\infty}$, so once again we must use l’Hôpital’s rule. This time we apply it twice in succession (because I’m not sure whether or not you can just cancel out the $x$s or not since it’s technically dividing by infinity; ask your local calculus teacher).

$\displaystyle\lim_{x \to \infty} \dfrac{-8x}{4x} = - \dfrac{8}{4} = \boxed{-2}$

Now, we put these solved limits back into the original problem:

$\dfrac{\dfrac{1}{10} \cdot 2}{6 + (-2)} = \dfrac{\dfrac{1}{5}}{4} = \boxed{\dfrac{1}{20}}$

Andy? We both fail. XD;