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Math Help - Limit of a Trigonometric Function

  1. #1
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    Limit of a Trigonometric Function

    I need help taking the limit of this problem please:

    f(x) = sin(x) / [x + tan(x)] as x approaches 0

    Any help would be appreciated, thank you.
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    Quote Originally Posted by jmhogart View Post
    I need help taking the limit of this problem please:

    f(x) = sin(x) / x + tan(x) as x approaches 0

    Any help would be appreciated, thank you.
    For this, you should be familiar with the special limit \lim_{x\to0}\frac{\sin x}x = 1\text.
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  3. #3
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    I am familiar with that case... but I am unsure of how to apply it. The answer that was shown was 1/2 and I am at a loss for how to get to that.
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    Quote Originally Posted by jmhogart View Post
    I need help taking the limit of this problem please:

    f(x) = sin(x) / [x + tan(x)] as x approaches 0

    Any help would be appreciated, thank you.
    Thankfully the edit you just did has removed a confusing ambiguity.

    The lazy way is to use l'Hopitals Rule. Have you been taught it?

    Alternatively, you can substitute the Maclaurin Series for sin x and tan x, then cancel out the common factor of x in the numerator and denominator and then take the limit without trouble.
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  5. #5
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    Yea sorry about that, I just realized it. I have not been taught that rule yet. We just started taking limits and derivatives of trigonometric functions and we are supposed to use the trigonometric identities to solve that problem.
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    Quote Originally Posted by jmhogart View Post
    Yea sorry about that, I just realized it. I have not been taught that rule yet. We just started taking limits and derivatives of trigonometric functions and we are supposed to use the trigonometric identities to solve that problem.
    In that case:

    \lim_{x\to0}\frac{\sin x}{x+\tan x}=\lim_{x\to0}\frac{\sin x}{x[1+(\tan x)/x]}=\lim_{x\to0}\left(\frac{\sin x}x\right)\left(\frac1{1+(\tan x)/x}\right)

    =\lim_{x\to0}\left(\frac{\sin x}x\right)\lim_{x\to0}\frac1{1+(\sin x)/(x\cos x)}

    =1\cdot\lim_{x\to0}\frac1{1+[(\sin x)/x][1/\cos x]}

    =\frac1{1+[\lim_{x\to0}(\sin x)/x][\lim_{x\to0}(1/\cos x)]}

    =\frac1{1+1\cdot1} = \frac12
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  7. #7
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    THANK YOU VERY MUCH!!! That never even crossed my mind, I was quite lost on what my first step should even be. I appreciate the help, once again thank you very much!!
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