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Math Help - Limit of Trigonometric function

  1. #1
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    Limit of Trigonometric function

    I came into a few problems in which requires me to take limits of Trigonometric function. Now I have no idea how to start, but below is a couple of the problems.

    1. Lim sin4x/8x
    x->0

    2. Lim tan6x/sin2x
    x->0

    I remember the rules for taking derivatives of Trigonometric functions, but I did not learn the foundation of these rules. So if anyone can help, please explain to me how to take limit of sin, cos, and tan.

    Thank you very much!
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  2. #2
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    One useful rule in this content is L'Hopital's rule: if differentiable functions f and g both vanish at x=a, but the derivative g' does not, then \frac{f}{g} \rightarrow \frac{f'(a)}{g'(a)}.
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  3. #3
    TD!
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    Although L'Hopital works fine, these problems look like they have to be done without L'Hopital, by using a default limit you've probably seen, which is:

    \mathop {\lim }\limits_{\alpha  \to 0} \frac{{\sin \alpha }}{\alpha } = 1

    For example, rewrite the first one as:

    \mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{8x}} = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{4x}}
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  4. #4
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    Thanks, but I understand that I have to simplify them.

    But for the second problem, I understand that

    Lim (x->0) 1/cos x = 1

    therefore I simplify (tan6x)/(sin2x) into (sin6x)/[(cos6x)*(sin2x)]

    Take the limit (x->0) out of it, the answer is 1/2

    Is that correct?
    Last edited by tttcomrader; April 2nd 2006 at 09:21 AM.
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  5. #5
    TD!
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    Watch out, in the standard limit I gave, the angle has to be the same as the lineair factor. So rewrite:

    <br />
\frac{{\tan \left( {6x} \right)}}{{\sin \left( {2x} \right)}} = \frac{{\sin \left( {6x} \right)}}{{\sin \left( {2x} \right)\cos \left( {6x} \right)}} = \frac{6}{2}\left( {\frac{{\sin \left( {6x} \right)}}{6}\frac{2}{{\sin \left( {2x} \right)}}\frac{1}{{\cos \left( {2x} \right)}}} \right)<br />

    And then take the limit

    <br />
3\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin \left( {6x} \right)}}{6}\frac{2}{{\sin \left( {2x} \right)}}\frac{1}{{\cos \left( {2x} \right)}}} \right) = 3 \cdot 1 \cdot 1 \cdot 1 = 3<br />
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  6. #6
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    Quote Originally Posted by TD!

    <br />
3\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin \left( {6x} \right)}}{6}\frac{2}{{\sin \left( {2x} \right)}}\frac{1}{{\cos \left( {2x} \right)}}} \right) = 3 \cdot 1 \cdot 1 \cdot 1 = 3<br />
    WHAT?!?
    The limit of \frac{\sin 6x}{6}=1?
    You probably want to say, \frac{\sin 6x}{6x}=1 and you seem to forgot to state that.
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  7. #7
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    Sorry but I"m getting a bit confused.

    Where does the 2 from the 2/[sin(2x)] and the 6 from (tan6x)/6x come from?
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  8. #8
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    Quote Originally Posted by tttcomrader
    Sorry but I"m getting a bit confused.

    Where does the 2 from the 2/[sin(2x)] and the 6 from (tan6x)/6x come from?
    He's multiplying and dividing by 2. Quick example:
    sin(2x)=\frac{2sin(2x)}{2}=2 \left ( \frac{sin(2x)}{2} \right ).

    Note the extra factors 6/2 on the outside of TD's expression.

    (And, of course, ThePerfectHacker is correct, he really needed to be putting the x's in as well.)

    -Dan
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  9. #9
    TD!
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    I left out the x's to see if you were all paying attention


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