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Solving limits containing trig expressions

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{x\rightarrow \frac{\pi}{2}} \frac{tan(2x)}{x - \frac{\pi}{2}}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I was given a couple of hints: use substitution, and that there isn't any need for the tangent double angle formula.

    I would have never thought to use substitution if I tried to solve this all day, and I had been trying to manipulate the tangent double angle formula for like an hour before I was told I didn't need it.

    How do you know when to use methods like substitution and when not to use the double angle formulas? Especially when there are like 30 of these on an exam, I can't even solve this 1 problem within an hour and I have to solve 30 in 45 minutes.

    Anyway, I got (actually I didn't, since I was told to do all of this and could never have though of any of this on my own):

    [tex]\lim_{x\rightarrow (x - h)} \frac{sin(2h + \pi)}{h \cdot cos(2h + \pi)} \\
    \lim_{x\rightarrow (x - h)} \frac{-sin(2h)}{-h \cdot cos(2h)} \\
    \lim_{x\rightarrow (x - h)} \frac{sin(2h)}{h \cdot cos(2h)}[/tex]

    I don't know what I should do now. Should I convert back to tangent? Use double angle formulas? More substitution? Something else? How do you know what exactly to do at this point?
     
    Last edited: Aug 29, 2012
  2. jcsd
  3. Aug 29, 2012 #2

    Mark44

    Staff: Mentor

    The LaTeX tags are case sensitive - use [ tex ] and [/ tex ] (no spaces), not [TEX] and [/TEX].
     
    Last edited: Aug 29, 2012
  4. Aug 29, 2012 #3

    Mark44

    Staff: Mentor

    I fixed it in my post.
     
  5. Aug 29, 2012 #4

    vela

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    Staff Emeritus
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    You do a bunch of problems and learn what works best in various cases. You'll develop your intuition over time.

    Also, you need to understand what the notation means. You need to get the basic stuff like that down otherwise you'll just make learning the rest of the material more difficult.

    The denominator goes to 0 in the original problem, so sometimes it helps to rewrite the limit in terms of a variable going to 0. So what you did was let ##h = x-\pi/2##. Then when you rewrite the problem in terms of h, you get
    $$\lim_{h\to 0} \frac{\sin 2h}{h\cos 2h}.$$ At this point, look up some of the special limits you should have learned about in class and see if you can see how they might help you in evaluating this one.
     
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