Limits

September 26th, 2009 by ben Leave a reply »

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;

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2 comments

  1. k says:

    Hey, I think I know the answer to this one. 1/20?

  2. Benji says:

    Good job!! You’re a genius.

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