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Math Help - How to solve limits?

  1. #1
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    Question How to solve limits?

    1. Limit [x->1] [(e^x-e)*(x-1)] = e

    How to solve ? Derivation (L'Hopital's rule) don't get anything useful - it remains [0/0].

    2. Limit [x->0] [Ln[Cos[x]] / x^2] = -1/2

    Any idea ?

    3. Limit [x->Inf] [(x^3+2x-1)^(1/3) / (x+2)] = 1

    How to solve ?

    4. [a^b]' = b*Ln[a]*[a]'
    Is it right?
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  2. #2
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    \displaystyle \lim_{x\to 1}\frac{e^x-e}{x-1}=\frac{0}{0}\Rightarrow \lim_{x\to 1}\frac{e^x}{1}=e
    I think you need a division sign not multiplication.

    \displaystyle \lim_{x\to 0}\frac{0}{0}\Rightarrow \lim_{x\to 0}\frac{\frac{-sin(x)}{cos(x)}}{2x}=\frac{0}{0}\Rightarrow\lim_{x  \to 0}\frac{-sec^2(x)}{2}=\frac{-1}{2}
    L'Hopital's Rule twice.

    \displaystyle \lim_{x\to \infty}\frac{(x^3)^{1/3}}{x}=\lim_{x\to \infty}\frac{x}{x}=1
    Limits to infinity isolate the highest powered exponents in the numerator and denominator.

    What is going on for number 4?
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  3. #3
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    dwsmith:
    How? If we apply L'Hopital's rule = > (-e+x*e^x)/(x-1)^2 => (-e+e)/(1-1)^2 => [0/0] ???
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  4. #4
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    e is constant so the derivative of e is 0 and the same for 1.

    The derivative of e^x is e^x and the derivative of x is 1.
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  5. #5
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    Quote Originally Posted by RCola View Post
    1. Limit [x->1] [(e^x-e)*(x-1)] = e
    here is what you originally posted ...

    \displaystyle \lim_{x \to 1} (e^x-e)(x-1)

    this limit is 0 ... what limit expression did you really mean to post?
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  6. #6
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    Quote Originally Posted by skeeter View Post
    here is what you originally posted ...

    \displaystyle \lim_{x \to 1} (e^x-e)(x-1)

    this limit is 0 ... what limit expression did you really mean to post?
    Division.
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  7. #7
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    Quote Originally Posted by dwsmith View Post
    e is constant so the derivative of e is 0 and the same for 1.

    The derivative of e^x is e^x and the derivative of x is 1.
    Aren't we suppose to derivate whole function instead of separately derivating numerator and denominator ?

    Derivating whole function I've got :

    [(e^x-e)/(x-1)^1]' = (e^x)*(x-1)^(-1) + (e^x-e)(-(x-1)^(-2)) = (e^x(x-1) - e^x+e ) / (x-1)^2 = e^x / (x-1)

    So : lim[x->1, (e^x)/(x-1)] = e/0 => we aren't allowd to devide by zero => deadlock

    Have I somewhere got mistake ?
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  8. #8
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    No, derivative of the numerator and denominator separately.
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  9. #9
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    Quote Originally Posted by skeeter View Post
    here is what you originally posted ...

    \displaystyle \lim_{x \to 1} (e^x-e)(x-1)

    this limit is 0 ... what limit expression did you really mean to post?
    Srry, It has to be (e^x-e)/(x-1)
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  10. #10
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    Quote Originally Posted by dwsmith View Post
    No, derivative of the numerator and denominator separately.
    Oh, I see, Lim [f(a)/f(b)] = lim [f'(a)/f'(b)]
    thx
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  11. #11
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    Quote Originally Posted by RCola View Post
    Oh, I see, Lim [f(a)/f(b)] = lim [f'(a)/f'(b)]
    thx
    Only if the function tends to \displaystyle \frac{0}{0} or \displaystyle \frac{\infty}{\infty}.
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  12. #12
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    Quote Originally Posted by dwsmith View Post
    What is going on for number 4?
    is it allowed to use this technique to find derivative ?
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  13. #13
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    Derivative with respect to what? a, b, x? Are a and b Real numbers?
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  14. #14
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    Quote Originally Posted by dwsmith View Post
    Derivative with respect to what? a, b, x? Are a and b Real numbers?
    with respect to x
    a and b is Real numbers
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  15. #15
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    Derivative is 0 then.
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