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Math Help - Limits of Indeterminate Forms and Improper Integrals

  1. #1
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    Limits of Indeterminate Forms and Improper Integrals

    Hey guys, I have a few problems I need help with and answers checked.

    Problem 1. Calculate each limit.

    a. \lim_{x \to 1}\frac{1 - x}{e^{x^3} - xe^x}

    My Solution

    Form \frac{0}{0}, Using L'Hopital's Rule. = \lim_{x \to 1}\frac{-1}{3x^{2}e^{x^3} - e^x + xe^x} = \frac{-e}{3}

    b. \lim_{x \to 0^+}\frac{e^{x^2} - 1}{\sin{x} - x}

    My Solution

    Form \frac{0}{0}, Using L'Hopital's Rule. = \lim_{x \to 0^+}\frac{2xe^{x^2}}{\cos{x} - 1}  = \frac{0}{0}, Using L'Hopital's Rule. = \lim_{x \to 0^+}\frac{2e^{x^2} + 4x^{2}e^{x^2}}{-\sin{x}} = -\infty

    c.
    \lim_{x \to 0}(\frac{1}{e^x - 1} - \frac{1}{x})

    My Solution

    I'm having trouble with this one.

    = \lim_{x \to 0}\frac{x - e^x + 1}{xe^x - x} = Form \frac{0}{0}, Using L'Hopital's Rule. = \lim_{x \to 0}\frac{-2x + x^{2}e^x - e^{2x} - xe^{2x} + 2e^x - 1}{x^{2}e^{2x} - 2x^{2}e^x + x^2} ... This is also of the form \frac{0}{0}. I've applied L'Hopital's Rule two more times and it just keeps getting nastier and nastier but is still keeping the form \frac{0}{0}. Where am I going wrong?

    d. \lim_{x \to \infty}(1 + \frac{1}{x})^x

    My Solution

    Let y = (1 + \frac{1}{x})^x, ln y = ln(1 + \frac{1}{x})^x = xln(1 + \frac{1}{x})

    \lim_{x \to \infty}ln y = \lim_{x \to \infty}xln(1 + \frac{1}{x}) = \lim_{x \to \infty}\frac{ln(1 + \frac{1}{x})}{\frac{1}{x}} = Form \frac{0}{0}, Using L'Hopital's Rule. = (Shortening the simplifying steps due to me not wanting carpal tunnel syndrome.) \lim_{x \to \infty}\frac{\frac{\frac{-1}{x^3}}{1 + \frac{1}{x}} - ln(1 + \frac{1}{x})(\frac{-1}{x^2})}{\frac{1}{x^2}} = \lim_{x \to \infty}\frac{-x^4 + x^{2}ln(1 + \frac{1}{x})(x^3 + x^2)}{x^5 + x^4} = -1, lny \to -1, y \to e^{-1}

    Problem 2. Determine whether the improper integral converges, or diverges. In the case of convergence, give its value.

    a.
    \int^{e}_{0}lnx dx

    My Solution


    = \lim_{t \to 0^+}\int^{e}_{t} lnx dx

    u = lnx, du = \frac{1}{x} dx, dv = dx, v = x

    = \lim_{t \to 0^+}xlnx - \int^{e}_{t} dx

    Here I don't know what I'm supposed to do. How am I supposed to get the limit if I have both x's and t's?

    b. \int^{\infty}_{2}xe^{-x} dx

    My Solution

    Here I run into the same problem as in 2a.

    Any and all help would be much appreciated. Thanks in advance.
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  2. #2
    Super Member Random Variable's Avatar
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     \lim_{b \to \infty} \int^{b}_{2} xe^{-x} \ dx = \lim_{b \to \infty} -xe^{-x} \Big|^{b}_{2} + \lim_{b \to \infty} \int^{b}_{2} e^{-x} \ dx

     =  \lim_{b \to \infty} (-xe^{-x} - e^{-x} \big|^{b}_2)

     = \lim_{b \to \infty} (-be^{-b} - e^{-b} +2e^{-2}+e^{-2}) = 3e^{-2}

    because  \lim_{b \to \infty} be^{-b} = \lim_{b \to \infty} \frac{b}{e^{b}} = \lim_{b \to \infty} \frac{1}{e^{b}} = 0
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  3. #3
    Super Member Random Variable's Avatar
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     \lim_{t \to 0+} \int^{e}_{t} \ln x \ dx = \lim_{t \to 0+} (x \ln x -x )\Big|^{e}_{t}

     = \lim_{t \to 0+}(e \ln e - e - t \ln t + t)

     = \lim_{t \to 0+}(e  - e - t \ln t + t) = 0

    because  \lim_{t \to 0+} t \ln t = \lim_{t \to 0+} \frac{ \ln t}{1/t}= \frac{1/t}{-1/t^{2}} = \lim_{t \to 0+} -t = 0
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  4. #4
    Super Member Random Variable's Avatar
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    1a) It should be  \lim_{x \to 1}\frac{-1}{3x^{2}e^{x^3} - e^x \textcolor{red}{-} xe^x} .


    1c) I'm sure what you did.

     \lim_{x \to 0} \frac{x-e^{x}+1}{xe^{x}-x} = \lim_{x \to 0} \frac{1-e^{x}}{e^{x}+xe^{x}-1}=
     = \lim_{x \to 0}  \frac{-e^{x}}{e^{x}+e^{x}+xe^{x}}= -\frac{1}{2}


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  5. #5
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    Thank you very much for your help. Did I do problems 1b and 1c right?
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  6. #6
    Super Member Aryth's Avatar
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    As far as I can tell, b is correct. For c I'll start with your consolidated form of it:

    \lim_{x \to 0} \frac{x - e^x + 1}{xe^x - x}

    Using L'Hopital's rule, we must differentiate the function:

    \frac{1 - e^x}{(e^x + xe^x) - 1}

    This is because we must differentiate the numerator and denominator separately, so that:

    \lim_{x \to 0} \frac{1 - e^x}{(e^x + xe^x) - 1} = \frac{0}{0}

    So, we use L'Hopital's rule again, getting:

    \frac{-e^x}{e^x + (e^x + xe^x)}

    So that:

    \lim_{x \to 0} \frac{-e^x}{e^x(2 + x)} = -\lim_{x \to 0} \frac{1}{2 + x} = -\frac{1}{2}

    So, Random Variable is right on b.
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  7. #7
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    I'm sorry, I meant is my solution to 1d correct (not 1b)?
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