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Math Help - L'Hospital's Rule

  1. #1
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    L'Hospital's Rule

    use l'hospital's rule to find the limit:

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

    i got 1 using ln y, but im not sure if its right..

    2. lim (1+ (a/x))^bx
    x-> INFINITY

    again i got 1...but am unsure

    any help is appreciated...thanks!!
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  2. #2
    Super Member 11rdc11's Avatar
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    I think the second one is the number e but just looked at it real fast
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  3. #3
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    Hello,
    Quote Originally Posted by skabani View Post
    use l'hospital's rule to find the limit:

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

    i got 1 using ln y, but im not sure if its right..
    I get 1 too

    2. lim (1+ (a/x))^bx
    x-> INFINITY
    Quote Originally Posted by 11rdc11 View Post
    I think the second one is the number e but just looked at it real fast
    Substitute t=\frac xa
    Thus bx=abt and \frac ax=\frac 1t

    \lim_{x \to \infty} \left(1+\frac ax\right)^{bx}=\lim_{t \to \infty} \left(1+\frac 1t\right)^{t \cdot (ab)}

    Remember the rule (a^b)^c=(a^c)^b=a^{bc}

    Therefore :

    \lim_{t \to \infty} \left(1+\frac 1t\right)^{t \cdot (ab)}=\lim_{t \to +\infty} \left(\left(1+\frac 1t\right)^t\right)^{ab}=e^{ab} \quad (*)

    (*) \lim_{t \to + \infty} \left(1+\frac 1t\right)^t=e (by taking the logarithm and applying l'Hospital's rule)
    (a little substitution can be done if t tends to -\infty)


    Otherwise, applying the logarithm :

    \ln(y)=\lim_{x \to \infty} bx \left(1+\frac ax\right)

    Then apply l'Hospital's rule...
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