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Math Help - Trigonometric limits - question involving tan.

  1. #1
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    Trigonometric limits - question involving tan.

    I am having trouble answering this question:

    "Using  tanx = \frac{sinx}{cosx} evaluate *limit as x approaches 0*  \frac{tanx - sinx}{xcosx} "

    My main problem, I think, is getting rid of the x on the bottom, I do not know how to go about doing it.

    My solution so far:
    note:*the "limit as x approaches 0" I don't know how to input on latex, so pretend it is there before all these.*

     = \frac{\frac{sinx}{cosx} - sinx}{xcosx}
     = \frac{\frac{sinx - sinxcosx}{cosx}}{xcosx}
     = \frac {sinx - sinx}{xcosx}

    Don't know what to do from here. I know the answer is zero, but I do not know how to get there from what I have. Help please.
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  3. #3
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    Hello, Kakariki!

    We're expected to know: . \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1


    I am having trouble answering this question:

    Using  \tan x \:=\: \frac{\sin x}{\cos x}, evaluate: . \lim_{x\to0}\frac{\tan x - \sin x}{x\cos x}

    My main problem, I think, is getting rid of the x on the bottom, I do not know how to go about doing it.

    My solution so far:

    \lim_{x\to0} \frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x}

    . . =\;\lim_{x\to0}\frac{\frac{\sin x - \sin x\cos x}{\cos x}}{x\cos x}

    . . = \;\lim_{x\to0}\frac {\sin x - \sin x}{x\cos x} . . . . wrong!

    We have: . \frac{\tan x - \sin x}{x\cos x} \;=\;\frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x}

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


    \text{Take the limit: }\;\lim_{x\to0}\bigg[\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}\bigg] \;=\;\underbrace{\lim_{x\to0}\left(\frac{\sin x}{x}\right)}_1\cdot\underbrace{\lim_{x\to0}\left(  \frac{1-\cos x}{\cos^2\!x}\right)}_{\frac{1-1}{1^2}}

    . . . . . . . . . = \;1\cdot0 \;=\;0

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Kakariki!

    We're expected to know: . \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1



    We have: . \frac{\tan x - \sin x}{x\cos x} \;=\;\frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x}

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


    \text{Take the limit: }\;\lim_{x\to0}\bigg[\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}\bigg] \;=\;\underbrace{\lim_{x\to0}\left(\frac{\sin x}{x}\right)}_1\cdot\underbrace{\lim_{x\to0}\left(  \frac{1-\cos x}{\cos^2\!x}\right)}_{\frac{1-1}{1^2}}

    . . . . . . . . . = \;1\cdot0 \;=\;0

    Okay, that makes sense. How did you know to multiply by  \frac{cosx}{cosx} though? Is it because you can't have a 0 in the denominator, but you can in the numerator?
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