DQ=Dan Quayle

Filed on Today, at 8:41 am

So today, I swam a mile, did one-arm pushups, and played soccer. I think I beat Ubemaya. But I have simple limit problem:
\lim_{x\rightarrow3}\frac{x^{2}-4x+3}{x^{2}-9}

In Blog news, the shoutbox appears to be shot and crippled for life since it is moving so slow. I am adding pianoforte to the contribs (Even if pianoforte does nothing).

Clicky Clicky

5 Comments have been made

Lhopital SPAM to the rescue!

The limit is of the form \frac{0}{0}. We spam Lhopital on it. Derivative of top is 2x - 4.
Derivative of bottom is 2x. Plug in three and BAM the limit is equal to \frac{1}{3}

Good job.

I just did
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\frac{x^{2}-4x+3}{x^{2}-9}=\frac{(x-3)(x-1)}{(x-3)(x+3)}=\frac{x-1}{x+3}
\lim_{x\rightarrow3}\frac{x-1}{x+3}=\frac{2}{6}=\boxed{\frac{1}{3}}
  • Posted Wed Sep 26, 2007 8:07 pm, by hunter34

That's what I did. Much faster.
  • Posted Thu Sep 27, 2007 11:43 am, by xpmath

How is factoring faster than L'hopital's?!
  • Posted Thu Sep 27, 2007 1:20 pm, by Temperal

Yea, but what if you get an unfactorable numerator (say, a trig function?) and a factorable denominator... You can factor all you want but eventually it comes down to either the squeeze theorem or Lhopital.

With alot of help from serialk11r

hunter34

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  • Joined: 12 Feb 2007
  • Location: Dallas, Texas
  • Occupation: Student. And about to be fired.
  • Interests: Soccer, math, sleep, anime, manga, food, sparta.

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