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Math Help - Trigonometric limits?

  1. #1
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    Trigonometric limits?

    How Would I solve these two limit

    1. As x approaches zero

    Sin(x)^2/x

    2. As x approaches zero

    Sec(x)-1/x sec(x)

    This is what I have done

    ((1/cos(x)-1))/(x)(1/cos(x))
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  2. #2
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    Re: Trigonometric limits?

    Hello, homeylova223!

    By now, you should know that more parenthese are needed . . .



    \text{(1) as }x\text{ approaches zero: }\:Sin(x)^2/x

    Is that: . \frac{\sin x^2}{x}\;\text{ or }\;\frac{\sin^2\!x}{x}\;?




    \text{(2) As }x\text{ approaches zero: }\:Sec(x)-1/x sec(x)

    Is that: . \sec x - \frac{1}{x}\sec x\;\text{ or }\;\sec(x) - \frac{1}{x\sec x}\;\text{ or }\;\frac{\sec x - 1}{x\sec x}\;?

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    Re: Trigonometric limits?

    For the both it is the latter

    (sin^2(x))/(x) This one I've managed to solve using a graphic calculator but I am wondering how to do it analytically

    ((secx-1))/((x sec x))
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  4. #4
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    Re: Trigonometric limits?

    Quote Originally Posted by homeylova223 View Post
    For the both it is the latter

    (sin^2(x))/(x) This one I've managed to solve using a graphic calculator but I am wondering how to do it analytically

    ((secx-1))/((x sec x))
    For Q.1, write it as \displaystyle \lim_{x \to 0}\frac{\sin{x}}{x}\cdot \lim_{x \to 0}\sin{x}. The first is a standard limit, the second can be evaluated using direct substitution.

    For Q.2
    \displaystyle \begin{align*}\lim_{x \to 0}\frac{\sec{x} - 1}{x\sec{x}} &= \lim_{x \to 0}\frac{\frac{1-\cos{x}}{\cos{x}}}{\frac{x}{\cos{x}}} \\ &= \lim_{x \to 0}\frac{1 - \cos{x}}{x}\end{align*}

    This is another standard limit, but if you don't know it, try multiplying top and bottom by the top's conjugate.
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    Re: Trigonometric limits?

    Hmm For Q.2 I do not understand how you can go from

    ((1/cos(x)-1))/(x)(1/cos(x))

    to (1-cos(x)/(cos(x)/(x)/(cosx)
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  6. #6
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    Re: Trigonometric limits?

    Quote Originally Posted by Prove It View Post
    try multiplying top and bottom by the top's conjugate.
    Quote Originally Posted by homeylova223 View Post
    Hmm For Q.2 I do not understand how you can go from

    ((1/cos(x)-1))/(x)(1/cos(x))

    to (1-cos(x)/(cos(x)/(x)/(cosx)
    Taking Prove It's suggestion multiply by the top's conjugate

    \displaystyle \lim_{x \to 0} \dfrac{1-\cos(x)}{x} \cdot \dfrac{1+\cos(x)}{1+\cos(x)} = \lim_{x \to 0} \dfrac{1-\cos^2(x)}{x(1+\cos(x))}

    edit: correction
    Last edited by e^(i*pi); July 13th 2011 at 02:57 PM.
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  7. #7
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    Re: Trigonometric limits?

    Quote Originally Posted by e^(i*pi) View Post
    \displaystyle \lim_{x \to 0} \dfrac{1-\cos(x)}{x} \cdot \dfrac{1+\cos(x)}{1+\cos(x)} = \lim_{x \to 0} \dfrac{1-\cos^2(x)}{1+\cos(x)}
    Not quite, you forgot a term in the last expression.
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    Re: Trigonometric limits?

    Quote Originally Posted by homeylova223 View Post
    Hmm For Q.2 I do not understand how you can go from

    ((1/cos(x)-1))/(x)(1/cos(x))

    to (1-cos(x)/(cos(x)/(x)/(cosx)
    Get a common denominator for \displaystyle \frac{1}{\cos{x}} - 1
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