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L'Hopital's Lim x->0 tan x / x

  1. Apr 15, 2007 #1
    I've done this problem but am not comfortable with the answer. Could someone take a quick look?

    Limx->0 tan x /x

    Limx->0 sec^2x ==> 0

    I'm just not sure this is right...
     
  2. jcsd
  3. Apr 15, 2007 #2

    quasar987

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    sec(x)=1/cos(x)

    So how do you conclude that the limit of sec²(x) is 0 ?
     
  4. Apr 15, 2007 #3
    ok, so could I use sec^2(x) = 1/cos^2(x) then use the power reducing formula to make: limx->0 2/(1+cos2x)? Then as x->0, then the value would be 2/1 thus 2?
     
  5. Apr 15, 2007 #4

    quasar987

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    Do you think that cos(0)=0 ? That's sin's job.
     
  6. Apr 15, 2007 #5
    ahhh cos(0) = 1 and sin(0) = 0. I should have verified that b4... Given this, then 2/(1+cos2(0)) = 2/2 = 1. Knowing this, there are 2 things that I should have worked through: secx = 1/cosx and that cos(0) = 1.

    Thanks very much for your help!
     
  7. Apr 15, 2007 #6

    mathwonk

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    or just recall that sin/cos = tan and cos(0)= 1, so tan(x)/x has the same limit as sin(x)/x which everyone knows is 1.

    but still everything relies on knowing cos(0) =1.
     
  8. Apr 15, 2007 #7
    Thank you! I'm not quite sure I understand, but I get tan = sin/cos since tye hypenuses cancel but how can I prove tan(x)/x has the same limit as sin(x)/x? I need to read and understand some of the finer points, or I won't be able to answer some fundamental questions. Any suggestions as to where to start? I'm going to google it and try and find something. Having been away from this stuff for a very long time, it's tough to come back....

    Thanks for your info!
     
  9. Apr 15, 2007 #8

    quasar987

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    The keys to mathwonk's argument is that

    [tex]\frac{\tan(x)}{x}=\frac{1}{\cos(x)}\frac{\sin(x)}{x}[/tex]

    , that

    [tex]\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1[/tex]

    , and that

    [tex]\lim_{x\rightarrow a}f(x)g(x)=\lim_{x\rightarrow a}f(x)\cdot \lim_{x\rightarrow a}g(x)[/tex]

    (provided the last two limits exist).
     
  10. Apr 15, 2007 #9

    Gib Z

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    Have you learned Taylor Series yet? Since this is the limit as x goes to zero, the first term of tan x's taylor series Could have been substituted in and The limit would have been easy to see.
     
  11. Apr 15, 2007 #10
    I understand the first part since (opp/hyp)/(adj/hyp) thus opp/adj = tan. That part makes sense. I don't get sin(x)/x = 1. Is this just something I should accept? I understand sin^2+cos^2=1. Would that be part of it? As for the last part, that makes sense provided they both approach the limit from the same side.

    I found some articles on the web that sort of explain things, but I suspect that I'm going to have to read them a few times to really understand it!

    Thanks for your help!
     
  12. Apr 15, 2007 #11
    No, we have not reviewed Taylor Series yet. We have just started that section. Might there be an area in which I could review and get a jump on what to expect?

    THANKS!
     
  13. Apr 15, 2007 #12

    Gib Z

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  14. Apr 15, 2007 #13

    Gib Z

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    Quotes thing doesn't seem to want to work for me, But you get the point.
     
  15. Apr 16, 2007 #14
    I do get ur point: tan x = opp/adj :smile:

    As for point 2: limx->0 sin(x)/x = 1 vs. sin(x)/x = 1 ARE 2 different statements. Good point and I need to keep this in mind since they say very different things.

    I'll search now in google for 'Taylor Series'. IK Wikipedia has also been a source of good info for me too!

    Thanks!
     
  16. Apr 16, 2007 #15

    quasar987

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    Here is a proof: http://people.hofstra.edu/faculty/Stefan_Waner/trig/triglim.html

    If the inequality arguments are a little over your head, just remember the result because it's very useful. And come back to the proof to get your revenge when you've become stronger.
     
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