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Homework Help: Limits of trigonometric functions

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data

    lim x --->0 [tan(2+x)^3 - tan8]/x

    2. Relevant equations

    f'(a)=lim h-->0 f(a+h)-f(a)/h

    3. The attempt at a solution

    should i differentiate first? am i allowed to do that?
    Last edited: Jul 6, 2008
  2. jcsd
  3. Jul 6, 2008 #2
    h= x
    a+h=2+x=2+h so a= 2
    f'(x)=secx^2 * 3x
    f'(2)= sec2^2 * 6= 6sec4
  4. Jul 6, 2008 #3
    does this all look correct?
  5. Jul 6, 2008 #4
    What book are you using? Chapter and problem number?
  6. Jul 6, 2008 #5
    it a problem for a test, i just need to evaluate the limit above, this is what the teacher showed to do for a similar problem
  7. Jul 6, 2008 #6
    Ok then, you can help yourself! You're given a take home exam ... use your book.
  8. Jul 6, 2008 #7


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    Science Advisor

    First write the problem correctly. Is it
    [tex]\lim_{x\rightarrow 2}\frac{tan^3(2+x)- tan(8)}{x}[/tex]
    or is it
    [tex]\lim_{x\rightarrow 2}\frac{tan((2+x)^3)- tan(8)}{x}[/tex]

    If it is the second then it can be interpreted as the derivatve of tan(x3) at x= 2. What is that derivative? (it is NOT sec2(x^2) (3x)[. Use the chain rule.)
  9. Jul 6, 2008 #8
    the second one is right. but as x goes to 0
  10. Jul 6, 2008 #9
  11. Jul 6, 2008 #10
    i think that is right and then i just substitute in 2 right? so its
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