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Math Help - L'Hospital's rule

  1. #1
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    L'Hospital's rule

    Can anyone explain the procedures in solving these two problems using L'Hospital's rule?

    1. lim x-> 0+ (tan 2x)^x

    2. lim x-> 1 (2 - x)^tan((pi)x/2)
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  2. #2
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    If the function is of indeterminate form, then take the derivative and find the limit of that. Read up, a tad on L'Hop's rule, it is not a hard rule you just need to know when to apply it
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  3. #3
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    Hello, Eternal!

    Here's the second one . . .


    (2)\;\;\lim_{x\to1}\,(2 - x)^{\tan(\frac{\pi}{2}x)}

    Let: . y \:=\:(2-x)^{\tan(\frac{\pi}{2}x)}

    Take logs: . \ln(y) \;=\;\ln\left[(2-x)^{\tan(\frac{\pi}{2}x)}\right] \;=\;\tan(\tfrac{\pi}{2}x)\cdot\ln(2-x) \;=\;\frac{\ln(2-x)}{\cot(\frac{\pi}{2}x)} \quad\rightarrow\quad \frac{0}{0}

    Apply L'Hopital: . \ln(y) \;=\;\frac{\frac{-1}{2-x}} {-\frac{\pi}{2}\csc^2(\frac{\pi}{2}x)} \;=\;\frac{2\sin^2(\frac{\pi}{2}x)}{\pi(2-x)}


    Take limits: . \lim_{x\to1}\,\ln(y) \;=\;\lim_{x\to1}\,\frac{2\sin^2(\frac{\pi}{2}x)}{  \pi(2-x)} \;=\;\frac{2\cdot1^2}{\pi(1)} \;=\;\frac{2}{\pi}


    We have: . \ln(y) \:=\:\frac{2}{\pi}

    Therefore: . y \;=\;e^{\frac{2}{\pi}}


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  4. #4
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    Hello again, Eternal!

    It took a while to get the first one.
    I hope I'm right . . .


    (1)\;\;\lim_{x\to0^+}(\tan 2x)^x
    As given, the limit goes to 0^0 . . . an indeterminate form.


    Let: . y \:=\:(\tan2x)^x

    Take logs: . \ln(y) \:=\:\ln(\tan2x)^x \:=\;x\cdot\ln(\tan2x) \;=\;\frac{\ln(\tan2x)}{\frac{1}{x}} \;=\;\frac{\ln(\tan2x)}{x^{-1}} \quad \to \quad \frac{\text{-}\infty}{\infty}

    Apply L'Hopital: . \frac{\;\dfrac{2\sec^2\!2x}{\tan2x}\;}{-x^{-2}} \;=\;\frac{-2x^2\sec^2\!2x}{\tan2x} \;=\; \frac{-2x^2}{\sin2x\cos2x} \;=\;\frac{-4x^2}{2\sin2x\cos2x} \;=\;\frac{-4x^2}{\sin4x} \quad\to \quad \frac{0}{0}

    Apply L'Hopital: . \frac{-8x}{4\cos4x} \;=\;\frac{-2x}{\cos4x}

    Take the limit: . \lim_{x\to0^+}\frac{-2x}{\cos4x} \;=\;\frac{0}{1} \;=\;0


    We have: . \lim_{x\to0^+}\bigg[\ln(y)\bigg] \:=\:0,

    . . therefore: . \lim_{x\to0^+}\,y \;=\;e^0 \;=\;1

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