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Math Help - Limits

  1. #1
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    Limits

    Assuming the result lim x approaches 0 (sinx)/(x) = 1, evalute

    1) lim x approaches 0 (1-cosx)/(x)

    2) lim x approaches 0 (1-cosx)(x^2)

    3) lim x approaches 0 (tanx-sinx)/(x^3)

    Can anyone help, I would appreciate it. My trig is horrible.
    Last edited by sderosa518; November 28th 2009 at 10:23 AM.
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  2. #2
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    Quote Originally Posted by sderosa518 View Post
    Assuming the result lim x approaches 0 (sinx)/(x) = 1, evalute

    1) lim x approaches 0 (1-cosx)/(x)

    2) lim x approaches 0 (1-cosx)(x^2)

    3) lim x approaches 0 (tanx-sinx)/(x^3)

    Can anyone help, I would appreciate it. My trig is horrible.
    Remember that you can use l-hopitals rule for indeterminate forms. Have you tried that?

    Here's an example, number 1 should be easy enough for you if you apply l-hopitals rule, here is the second limit:

    \lim_{x->0}\frac{1-cos(x)}{x^2}=\lim_{x->0}\frac{sin(x)}{2x}

    =\frac{1}{2}\lim_{x->0}\frac{sin(x)}{x}=\frac{1}{2}

    As you can see, there isn't much trig involved in that one. It's just l-hopitals rule.
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  3. #3
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    You can use L'Hopital's rule. For (2) You can use the property of

    \lim_{x\to0}\frac{sin(x)}{x} = 1

    to solve for the limit after doing L'Hopital's rule.
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  4. #4
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    Hello, sderosa518!

    Assuming: . \lim_{x\to0}\frac{\sin x}{x} \:=\:1

    evaluate: . 1)\;\;\lim_{x\to0}\frac{1-\cos x}{x}

    We have: . \frac{1-\cos x}{x}

    Multiply by \frac{1+\cos x}{1 + \cos x}\!:\quad\frac{1-\cos x}{x}\cdot\frac{1+\cos x}{1 + \cos x} \;=\;\frac{1-\cos^2\!x}{x(1+\cos x)} \;=\;\frac{\sin^2\!x}{x(1+\cos x)} . = \;\frac{\sin x}{x}\cdot\frac{\sin x}{1 + \cos x}


    Then: . \lim_{x\to0}\left(\frac{\sin x}{x}\right)\left(\frac{\sin x}{1 + \cos x}\right) \;=\;(1)\left(\frac{0}{1+1}\right) \;=\;0




    2)\;\;\lim_{x\to0}\frac{1-\cos x}{x^2}

    Multiply by \frac{1+\cos x}{1+\cos x}\!:\quad \frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x} \;=\;\frac{1-\cos^2\!x}{x^2(1+\cos x)} . =\;\frac{\sin^2\!x}{x^2(1+\cos x)} \;=\;\frac{\sin^2\!x}{x^2}\cdot\frac{1}{1+\cos x}


    Then: . \lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\left(\frac{1}{1+\cos x}\right) \;=\;(1^2)\left(\frac{1}{1+1}\right) \;=\;\frac{1}{2}




    3)\;\;\lim_{x\to0}\frac{\tan x - \sin x}{x^3}

    We have: . \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \;=\;\frac{\sin x - \sin x\cos x}{x^3\cos x} \;=\;\frac{\sin x(1 - \cos x)}{x^3\cos x}

    Multiply by \frac{1+\cos x}{1+\cos x}\!:\quad \frac{\sin x(1 - \cos x)}{x^3\cos x}\cdot\frac{1+\cos x}{1 + \cos x} \;=\;\frac{\sin x(1-\cos^2\!x)}{x^3\cos x(1 + \cos x)}

    . . . . . . = \; \frac{\sin x\cdot\sin^2\!x}{x^3\cos x(1 + \cos x)} \;=\;\frac{\sin^3\!x}{x^3\cos x(1 + \cos x)} \;=\;\frac{\sin^3\!x}{x^3}\cdot\frac{1}{\cos x(1 + \cos x)}


    Then: . \lim_{x\to0}\left[\frac{\sin^3\!x}{x^3}\cdot\frac{1}{\cos x(1 + \cos x)}\right] \;=\;(1^3)\left(\frac{1}{1(1+1)}\right) \;=\;\frac{1}{2}

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