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Thread: L'Hopital's Questions

  1. #1
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    L'Hopital's Questions

    Hi

    I need some help on the following questions:
    1) Use L'Hopital's rule to obtain:$\displaystyle \lim_{x \to 0} \frac{e^{x}-x-1}{3x^2}$

    $\displaystyle f'(x) = \frac{3x^{2}*(e^{x}-1)-(e^{x}-x-1)*6x^2}{9x^4}$

    $\displaystyle f'(x) = \frac{3x^{2}e^{x} - 6xe^{x}+3x^2+6x}{9x^4}$

    $\displaystyle f'(x) = \frac{3x(xe^{x} - 2e^{x} +3x+2}{9x^4}$
    $\displaystyle
    f'(x) = \frac{(xe^{x} - 2e^{x} +3x+2}{x^3}$

    so if x=0, i get 0 however book's answer says $\displaystyle \frac{1}{6}$
    2) Use L'Hopital's rule to obtain: $\displaystyle \lim_{x \to \infty} \frac{x+2}{x^2+2x+1}$

    $\displaystyle f'(x) = \frac{1}{2x+2} = \frac{1}{2\infty+2}$


    The only problem i am having is what does $\displaystyle \infty$ equal to?

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi

    I need some help on the following questions:
    1) Use L'Hopital's rule to obtain:$\displaystyle \lim_{x \to 0} \frac{e^{x}-x-1}{3x^2}$

    $\displaystyle f'(x) = \frac{3x^{2}*(e^{x}-1)-(e^{x}-x-1)*6x^2}{9x^4}$

    $\displaystyle f'(x) = \frac{3x^{2}e^{x} - 6xe^{x}+3x^2+6x}{9x^4}$

    $\displaystyle f'(x) = \frac{3x(xe^{x} - 2e^{x} +3x+2}{9x^4}$
    $\displaystyle
    f'(x) = \frac{(xe^{x} - 2e^{x} +3x+2}{x^3}$

    so if x=0, i get 0 however book's answer says $\displaystyle \frac{1}{6}$
    This is NOT L'Hopital's rule! You have taken the limit of the derivative of f which is NOT, in general, the limit of f.
    (And even for that you would NOT get "0"- you would get $\displaystyle \frac{1}{0}$ which does not exist. Your f' does not have a limit.)

    L'Hopital's rule says that if, at x= a, $\displaystyle \frac{f(x)}{g(x)}$ is the indeterminant form "$\displaystyle \frac{0}{0}$" then $\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}= \frac{\lim_{x\to a} f'(x)}{\lim_{x\to a} g'(x)}$ provided the limits on the right exist.

    That is, you differentiate the numerator and denominator separately NOT using the "quotient law".
    Here, the numerator is $\displaystyle e^x- x- 1$ and the denominator is $\displaystyle 3x^2$ which are 0 when x= 0, giving the form $\displaystyle \frac{0}{0}$ so we can apply L'Hopital's rule.

    The derivative of $\displaystyle e^x- x- 1$ is $\displaystyle e^x- 1$ and the derivative of $\displaystyle 3x^2$ is $\displaystyle 6x$. Those are both 0 at x= 0 so apply L'Hopital's rule again.

    The derivative of $\displaystyle e^x- 1$ is $\displaystyle e^x$ and the derivative of $\displaystyle 6x$ is $\displaystyle 6$ which are 1 and 6 at x= 0.

    The limit is $\displaystyle \frac{1}{6}$.

    2) Use L'Hopital's rule to obtain: $\displaystyle \lim_{x \to \infty} \frac{x+2}{x^2+2x+1}$

    $\displaystyle f'(x) = \frac{1}{2x+2} = \frac{1}{2\infty+2}$
    Okay, here you did differentiate numerator and denominator separately- although it should not be labeled "f' ".


    The only problem i am having is what does $\displaystyle \infty$ equal to?

    P.S
    It doesn't equal anything- $\displaystyle \infty$ is not a number, just shorthand for "gets larger without bound."

    Here you have a fixed number, 1, over a denominator that gets larger and larger without bound- think of $\displaystyle \frac{1}{100}$, $\displaystyle \frac{1}{10000}$, $\displaystyle \frac{1}{10000000}$, etc. What is happening to those numbers as the denominator gets larger and larger?
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  3. #3
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    so for question 2 the book's answer says it 0 how did the get that??
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  4. #4
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    Did you look at the numbers I suggested?

    $\displaystyle \frac{1}{100}$, $\displaystyle \frac{1}{100000}$, $\displaystyle \frac{1}{100000000000}$ are equal to what in decimal form?

    If the numerator of a fraction is a constant, and the denominator gets larger and larger, what happens to the fraction?
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  5. #5
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    closer and closer to 0.
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