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Calculus 1: Limit

  1. Oct 6, 2008 #1
    So I was helping this person the other day, and this problem made me pretty mad.

    [tex]\lim_{x\rightarrow\frac{\pi}{9}}\left(\frac{\sin x}{x-\frac{\pi}{9}}\right)[/tex]

    First, I added and subtracted pi/9. I regrouped the angle so that it was sin[(x-pi/9)+pi/9] and expanded it, but that didn't help at all.

    I could use L'Hopital, but this is a calculus 1 problem.
  2. jcsd
  3. Oct 6, 2008 #2


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    Might have made you mad because the limit doesn't exist. sin(pi/9) is not equal to zero. The denominator does approach zero.
  4. Oct 7, 2008 #3
    Yeah, like Dick already pointed it out, this limit doesn't exist, and it is very easy to notice it by just observing some values to the left and to the right of [tex] \frac{\pi}{9}[/tex] since pi/9 lies in the first quadrant, sin(x) wont change sing.

    But the bottom, when we take values to the left of pi/9 will be negative, while to the right positive.
    SO the right hand side limit of this is infinity, while the left hand side -infinity. The overall limit does not exist.
  5. Oct 7, 2008 #4


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    And, therefore, you could not use L'Hopital's rule!
  6. Oct 7, 2008 #5
    oh shoot! This was a quiz problem and gave the answer -1. Now I'm mad!
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