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Trig Limit question

  1. Jul 5, 2009 #1
    Could someone help me with this one. My teacher says I'm wrong when I solved this limit as undefined. thanks!

    lim as x goes to zero of (sin x tan x -1)/4x
     
  2. jcsd
  3. Jul 6, 2009 #2
    If you mean (sin(x)tan(x)-1)/(4x), then it would be undefined. I have a feeling that the -1 is causing a problem since you may be missing parentheses that would change the problem and let a limit exist.
     
  4. Jul 6, 2009 #3
    If the limit you are asking about is
    [tex]\lim_{x\to 0} \frac{\sin x \tan x - 1}{4x}[/tex]
    then the limit is undefined, and your teacher is wrong. The way to determine this is to look at the limit as x approaches 0 from the left and then from the right.

    The limit of the numerator as x approaches 0 is -1. As x approaches 0 from the left, x is negative and so the limit will be positive infinity. As x approaches 0 from the right, x is positive, so you will get negative infinity. Therefore, the limit does not exist since the left and right handed limits aren't equal.
     
  5. Jul 6, 2009 #4
    thanks for the responses but when i graph this function at 0 the function is zero. It looks like it is continuous and goes to zero from the left and right but i can't find a mathematical way to solve it.
     
  6. Jul 6, 2009 #5

    Cyosis

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    Well you must have drawn the graph incorrectly. For example what value did you get for x=0.1 and x=-0.1?
     
  7. Jul 6, 2009 #6
    for -.01 the limit is .0025 and for .01 the limit is -.0025. looking at a graphing calculator using the table function, it looks like it converges to 0.
     
  8. Jul 6, 2009 #7

    Cyosis

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    I see, that means we're talking about different functions and so are the other posters in this thread.

    The limits you describe correspond with the function:

    [tex]
    x \left(\frac{\sin x \tan x -1}{4}\right)
    [/tex]

    We assumed you meant:

    [tex]
    \frac{\sin x \tan x - 1}{4x}
    [/tex]

    Could you clarify which one it is?
     
  9. Jul 6, 2009 #8
    my mistake on the calculator. it is definitely the second one. looking at the graph my teacher must be wrong (I hope!!!!!!) thanks for your help!!!!!!!!!
     
  10. Jul 6, 2009 #9

    Cyosis

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    If it's the second one then the limit does not exist.
     
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