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Math Help - evaluate limit

  1. #1
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    evaluate limit

    use L'Hopitals Rule
    as the limit x approaches 0 from the right side
    tanxlnx

    i got as far as this lnx/(1/tanx) and tried to differentiate but coudnt get anything that sounded right
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  2. #2
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    Hello, fgkl8!

    Use L'Hopital's Rule: . \lim_{x\to0^+} \tan x\ln x

    Since the limit goes to: . 0\cdot(\text{-}\infty) , we can apply L'Hopital.


    We have: . \tan x \ln x \:=\:\frac{\ln x}{\cot x}

    Apply L'Hopital: . \frac{\frac{1}{x}}{-\csc^2\!x} \;=\;-\frac{\sin^2\!x}{x} \;=\;-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)


    Therefore: . \lim_{x\to0^+}\bigg[-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)\bigg] \;=\;-(1)(\infty) \;=\;-\infty

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, fgkl8!


    Since the limit goes to: . 0\cdot(\text{-}\infty) , we can apply L'Hopital.


    We have: . \tan x \ln x \:=\:\frac{\ln x}{\cot x}

    Apply L'Hopital: . \frac{\frac{1}{x}}{-\csc^2\!x} \;=\;-\frac{\sin^2\!x}{x} \;=\;-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)

    correction ...

    Therefore: . \lim_{x\to0^+}\textcolor{red}{\bigg[-\left(\frac{\sin x}{x}\right)\left(\frac{\sin{x}}{1}\right)\bigg] \;=\;-(1)(0) = 0 }

    ...
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, fgkl8!


    Since the limit goes to: . 0\cdot(\text{-}\infty) , we can apply L'Hopital.


    We have: . \tan x \ln x \:=\:\frac{\ln x}{\cot x}

    Apply L'Hopital: . \frac{\frac{1}{x}}{-\csc^2\!x} \;=\;-\frac{\sin^2\!x}{x} \;=\;-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)
    EDIT: Was just a typo in. missed that a sin was missing just noticed the (sin/x)(1/x) bit.

    Therefore: . \lim_{x\to0^+}\bigg[-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)\bigg] \;=\;-(1)(\infty) \;=\;-\infty

    i dont agree with the last line, lim(ab)=lim (a)lim(b) if both limits exist.
    applying l'hopitals once again gives

    lim 2sinxcosx/1 as x goes to 0

    which is 0
    Last edited by jiboom; February 28th 2010 at 01:48 PM. Reason: EDIT: Was just a typo in soroban's post. missed that a sin was missing just noticed the (sin/x)(1/x) bit.
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  5. #5
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    for x\ne0 we have \tan (x)\ln (x)=\frac{\tan x}{x}\cdot x\ln (x), and we can compute its limit.
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