Finding Limit of Trig Func

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\lim_{x\to\0} \frac{2tan^2x}{x}[/itex]



    3. The attempt at a solution

    [itex]\lim_{x\to\0} \frac{2tan^2x}{x} \\
    = \frac{2 tanx tanx}{x} \\
    = \cfrac{2 \cfrac {sin}{cos} \cfrac {sin}{cos}[/itex]

    ?

    Edit: Oh boy none of my latex is working. :(

    1. The problem statement, all variables and given/known data
    lim x--> 0 (2tan^2x)/x


    3. The attempt at a solution

    lim x--> 0 (2tan^2x)/x
    = [ 2 tanx (tan x) ] / x
    = [ 2 (sin / cos) (sin/cos) ] / x
    = [ 2 (sin^2x/cos^2x) ] / x
    = [ 2sin^2x / cos^2x ] \ x

    Help please. Knowing me the answer is probably pretty simple. =D
     
  2. jcsd
  3. Oct 9, 2011 #2

    ehild

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    Do you know the limit x-->0 (sin(x)/x)?

    ehild
     
  4. Oct 9, 2011 #3
    Yes its equal to 1, couldn't fit it in here. But I did just get an idea at the moment. Here is what I did.

    lim x--> 0 (2tan^2x)/x
    = [ 2 tanx (tan x) ] / x
    = [ 2 (sin / cos) (sin/cos) ] / x
    From here on out, I decided to multiply the numerator and denominator by cos/sin
    = [ [ 2 (sin / cos) (sin/cos) ] / x ] cos/sin
    = [2 (cos/sin) ] / (cos/sin)
    = 2

    Can anyone confirm this? Its an even number in the book so I can't tell if I get it right or wrong, and cramster.com doesn't supply this problem.

    Also any ideas why my latex code didn't work?
     
  5. Oct 9, 2011 #4

    ehild

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    Your idea does not help and the result is wrong.


    ehild
     
  6. Oct 9, 2011 #5

    ehild

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    [tex]\lim_{x \to 0} \frac{tan^2(x)}{x}=\lim_{x \to 0} \frac{sin(x)}{x}\frac{sin(x)}{cos^2(x)}[/tex]

    ehild
     
  7. Oct 9, 2011 #6
    Do you know l'Hopital's rule?
     
  8. Oct 9, 2011 #7
    I'm having trouble following your logic.

    Nope. My test is only on Calculus I, I think that rule doesn't get introduced till later?
     
  9. Oct 9, 2011 #8
    How so, what did I do wrong?

    [tex] \lim_{x \to 0} \frac{2tan^2(x)}{x} = \frac{2tanx tan x }{x }
    = {2 \frac{sinx}{cosx} \frac{sin }{cos }
    \lim_{x \to 0}[/tex]

    Okay this is annoying, I don't know whats wrong with my latex code, I could swear I've writen everything the same style as you. I'm not gonna try to rewrite my whole idea in latex until I figure it out.

    What did I do wrong here?

    lim x--> 0 (2tan^2x)/x
    = [ 2 tanx (tan x) ] / x
    = [ 2 (sin / cos) (sin/cos) ] / x
    From here on out, I decided to multiply the numerator and denominator by cos/sin
    = [ [ 2 (sin / cos) (sin/cos) ] / x ] cos/sin
    = [2 (cos/sin) ] / (cos/sin)
    = 2

    I don't see that I broke any rule of algebra or misused one so please let me know.
     
    Last edited: Oct 9, 2011
  10. Oct 9, 2011 #9
    Can someone pleasee helppp?

    [tex]\lim_{x\to 0} \frac{2tan^2x}{x}[/tex]
    [tex]\lim_{x\to 0} =\frac{2 tan x (tan x)}{x}[/tex]
    [tex]\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x}[/tex]
    [tex]\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x} \frac{\frac{cos}{sin}}{\frac{cos}{sin}}[/tex]
    [tex]\lim_{x\to 0}= \frac{\frac{2cos}{sin}}{\frac{cosx}{sinx}}[/tex]
    [tex]\lim_{x\to 0}= \frac{2cos}{sin}\frac{sinx}{cosx}[/tex]
    [tex]\lim_{x\to 0}= \frac{2x}{x}[/tex]
    [tex]\lim_{x\to 0}= 2 [/tex]

    I took the time to put everything in clear latex form, I really want to figure this out. Input would be GREATLY appreciated!
     
    Last edited: Oct 9, 2011
  11. Oct 9, 2011 #10

    SammyS

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    Use the above !!!!

    limit of the product is the product of the limits.
     
  12. Oct 9, 2011 #11
    But I replied that I don't know the logic behind that step. It completely defeats the purpose if I don't know how he got to it.

    [tex]\lim_{x\to 0}\frac{tan2x}{x} = \lim_{x\to 0}\frac{\frac{sinx}{cosx}\frac{sinx}{cosx}}{x}[/tex]
    [tex]\lim_{x\to 0}=\frac{\frac {sinx}{1} \frac{sinx}{cos^2x}}{x}[/tex]
    [tex]\lim_{x\to 0}=(\frac{sinx}{1}\frac{sinx}{cos^2x})\frac{1}{x}[/tex]

    ?? So how did you go from [tex]\lim_{x\to 0}\frac{tan2x}{x}[/tex] to
    [tex]\lim_{x\to 0}\frac{sin(x)}{x}\frac{sin(x)}{cos^2(x)}[/tex]

    Even then I don't know how to solve the problem with statement above. But I first have to know how you get to that statement. Its not just about solving the problem, its very important to learn from the problem.
     
  13. Oct 9, 2011 #12

    SammyS

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    [itex]\displaystyle \tan^2(x)=\frac{\sin(x)}{\cos(x)}\frac{\sin(x)}{ \cos(x)}=\frac{\sin(x)}{1}\frac{\sin(x)}{\cos^2(x)}[/itex]

    ∴ [itex]\displaystyle \frac{\tan^2(x)}{x}=\frac{\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}[/itex]
     
  14. Oct 9, 2011 #13
    That is the same exact thing that I did BUT:

    [tex]\lim_{x\to 0}=\frac{\frac {sinx}{1} \frac{sinx}{cos^2x}}{x}[/tex]
    [tex]\lim_{x\to 0}=(\frac{sinx}{1}\frac{sinx}{cos^2x})\frac{1}{x}[/tex]
    I stopped here because wouldn't that actually equal:
    [tex]\lim_{x\to 0}=\frac {sinx}{x} \frac{sinx}{cos^2(x)^2}[/tex]
     
  15. Oct 9, 2011 #14

    SammyS

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    Almost, but:  [itex]x\,\cos(x)\ne\cos(x^2)[/itex]

    That is equal to [tex]\lim_{x\to 0}\left(\frac {\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}\right)=\lim_{x\to 0}\left(\frac {\sin(x)}{x}\right)\cdot \lim_{x\to 0}\left(\frac{\sin(x)}{\cos^2(x)}\right)[/tex]
     
  16. Oct 9, 2011 #15
    I suppose [itex]x\,\cos(x)\ne\cos(x^2)[/itex]
    because you can't multiply x by an angle?. What theorem shows that [tex](x) cos(x) = cos(x)[/tex] or [tex](x) cos^2 x = cos(x)[/tex]

    Besides that,

    [tex]=\lim_{x\to 0}\left(\frac {\sin(x)}{x}\right)\cdot \lim_{x\to 0}\left(\frac{\sin(x)}{\cos^2(x)}\right)[/tex]
    [tex]=1 \cdot \frac{0}{-1}[/tex]
    [tex]=0[/tex]

    ?

    *takes a deep breath*
     
  17. Oct 9, 2011 #16

    SammyS

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    Multiplication is not distributive with respect to multiplication.
     
  18. Oct 9, 2011 #17
    I'm the kind of person that feels very unsatisfied without a proof or a theorem to refer to. But I guess I'll stop pulling your rear.

    Thanks for your help, one final thing though; if this previous calculation was correct, could you please guide me what was wrong about the other calculation? I mean, I just don't understand what I did wrong it drives me crazy lol. How am I to get better if I don't learn from my mistakes? ^.^

    [tex]\lim_{x\to 0} \frac{2tan^2x}{x}[/tex]
    [tex]\lim_{x\to 0} =\frac{2 tan x (tan x)}{x}[/tex]
    [tex]\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x}[/tex]
    [tex]\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x} \frac{\frac{cos}{sin}}{\frac{cos}{sin}}[/tex]
    [tex]\lim_{x\to 0}= \frac{\frac{2cos}{sin}}{\frac{cosx}{sinx}}[/tex]
    [tex]\lim_{x\to 0}= \frac{2cos}{sin}\frac{sinx}{cosx}[/tex]
    [tex]\lim_{x\to 0}= \frac{2x}{x}[/tex]
    [tex]\lim_{x\to 0}= 2 [/tex]
     
  19. Oct 9, 2011 #18
    Personally I would just use l'Hopital's rule.
     
  20. Oct 10, 2011 #19
    Well I don't learn that till calculus 2, I looked it up real quick and it looks like a pretty straightforward theorem, but problem is I don't know how to take the derivative of the trigonometric function just yet.

    My class started a whopping (almost) 2 weeks late relative to other classes.
     
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