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Evaluating this limit

  1. Apr 15, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{h\rightarrow0}= \frac{tan(\frac{\pi}{4}+2h)\ - tan\frac{\pi}{4}}h[/tex]

    3. The attempt at a solution

    Here's what i was able to work out.

    Use the trig addition formula:

    [tex]\lim_{h\rightarrow0}= \frac{(\frac{tan\frac{\pi}{4}+ tan2h}{1-tan\frac{\pi}{4}tan2h}) - tan\frac{\pi}{4}}h[/tex]

    Get the LCD of 1 - tan(pi/4)tan2h, then multiply by 1 - tan(pi/4)tan2h to get rid of the fraction.

    [tex]\lim_{h\rightarrow0} = \frac{tan\frac{\pi}{4} + tan2h - tan\frac{\pi}{4}(1 - tan\frac{\pi}{4}tan2h)} {h(1- tan\frac{\pi}{4}tan2h)}[/tex]

    The last two terms cancel out:

    [tex]\lim_{h\rightarrow0} = \frac{tan\frac{\pi}{4} + tan2h - tan\frac{\pi}{4}}h[/tex]

    You're left with:

    [tex]\lim_{h\rightarrow0} = \frac{tan2h}h[/tex]

    Then i believe you would need to use the double angle formula?

    [tex]\lim_{h\rightarrow0} = \frac{\frac{2tanh}{1-tan^2h}}h[/tex]

    multiply to get rid of fraction

    [tex]\lim_{h\rightarrow0} = \frac{2tanh}{h(1 - tan^2h)}[/tex]

    And this is where I got stuck, any suggestions?
  2. jcsd
  3. Apr 15, 2008 #2
    sin(h)/h -- > 1 as h-->0 ** [edit Thanks]

    Can also use L'Hospital's Rule.
    Taylor Series ..
    Last edited: Apr 15, 2008
  4. Apr 15, 2008 #3
    Do you have to stick with canceling stuff out algebraically?

    If not just L'Hopital that sucker.
    You should get 2.
  5. Apr 15, 2008 #4

    Whoa, wait, what happened to the rest of the denominator, and other term in the numerator?
  6. Apr 15, 2008 #5
    sin(h)/h =/ ->1 as h ->1.
    lim as h-> 1 of sin(h)/h = sin1

    I think you're mixing it up with that one fact lim as x -> 0 of sinx/x = 1 or something like that.
  7. Apr 15, 2008 #6
    Using L'Hospital here would be putting the cart before the horse, since it requires already knowing what the derivative of tan() is, no?
  8. Apr 15, 2008 #7
    Can't use l'hopitals rule since the problem is given as a limit h--> 0. They don't give f(x)

    "Whoa, wait, what happened to the rest of the denominator, and other term in the numerator?"

    They cancel out, don't they?
  9. Apr 15, 2008 #8
    Why would that stop you?

  10. Apr 15, 2008 #9
    yeah, I was just kidding about that whole "2" business. The answer is actually 4. And I double checked myself this time.

    I'm assuming that you were taught that lim as h->0 of f(x) is the format you're used to, right?
    Here f(x) is not explicitly defined as such but the the fraction which you a taking the limit of is itself a function of the variable h so the same rules and concepts will apply. And since you get the case 0/0 you can go right ahead and l'hopital it.
    and to qudraphonics: do many people learn l'hopital's rule before they learn the derivative of tan(u)?
  11. Apr 15, 2008 #10
    I learned the derivative of tan(u) before I learned L'Hospital's rule, IIRC. But the point is that the limit in question is the definition of the derivative of tan(u) (well, up to that factor of 2), so it's sort of circular to suggest using L'Hospital's rule, which gives the answer in terms of the derivative of tan(u). I suppose you could apply enough trig identities so that it's no longer expressed in terms of tan(), and then apply L'Hospital to that (assuming you already know the derivative of the various trig functions in the expression).

    However, now it occurs to me that maybe this problem is aimed at people who already have learned all of the derivatives in question, as well as L'Hospital's rule, and they're just checking that you understand the effect of including the extra factor of 2 ("extra" compared to the definition of the derivative). In which case, L'Hospital away!
  12. Apr 16, 2008 #11


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    The point people are trying to make is that tan(2h)= sin(2h)/cos(2h).
    [tex]\frac{tan(2h)}{h}= \frac{sin(2h)}{h}\frac{1}{cos(2h)}= 2\frac{sin(2h)}{2h}\frac{1}{cos(2h)}[/tex]
    Now, what is the limit of that?
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