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Math Help - L'Hospital's Rule Example with e

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    L'Hospital's Rule Example with e

    \lim x\rightarrow 0[\dfrac{e^{9x} - 1 - 9x}{x^{2}}]


    \lim x\rightarrow 0[\dfrac{e^{9(0)} - 1 - 9(0)}{(0)^{2}}]


    \lim x\rightarrow 0[\dfrac{e^{0} - 1 - 0}{0}]


    \lim x\rightarrow 0[\dfrac{1 - 1 - 0}{0} = \dfrac{0}{0}] - Indeterminate

    \lim x\rightarrow 0[\dfrac{[(x^{2})(9e^{9x} - 9)] - [(e^{9x} - 1 - 9x)(2x)]}{[(x^{2})^{2}]} ] - Using L' Hospital's rule by way of quotient rule

    \lim x\rightarrow 0[\dfrac{[x^{2}9e^{9x}-9x^{2}] - [(2xe^{9x} - 2x - 18x)(2x)]}{x^{4}} Algebra right so far?
    Last edited by Jason76; October 24th 2013 at 08:19 PM.
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    Re: L'Hospital's Rule Example with e

    Quote Originally Posted by Jason76 View Post
    \lim x\rightarrow 0[\dfrac{e^{9x} - 1 - 9x}{x^{2}}]


    \lim x\rightarrow 0[\dfrac{e^{9(0)} - 1 - 9(0)}{(0)^{2}}]


    \lim x\rightarrow 0[\dfrac{e^{0} - 1 - 0}{0}]


    \lim x\rightarrow 0[\dfrac{1 - 1 - 0}{0} = \dfrac{0}{0}] - Indeterminate

    \lim x\rightarrow 0[\dfrac{[(x^{2})(9e^{9x} - 9)] - [(e^{9x} - 1 - 9x)(2x)]}{[(x^{2})^{2}]} ] - Using L' Hospital's rule by way of quotient rule
    You don't differentiate the entire function to use L'Hospital's Rule, you differentiate just the top and bottom. If the limit is indeterminate of the form \displaystyle \begin{align*} \frac{0}{0} \end{align*} or \displaystyle \begin{align*} \frac{\infty}{\infty}  \end{align*}, then \displaystyle \begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \end{align*}.
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    Re: L'Hospital's Rule Example with e

    Quote Originally Posted by Prove It View Post
    You don't differentiate the entire function to use L'Hospital's Rule, you differentiate just the top and bottom. If the limit is indeterminate of the form \displaystyle \begin{align*} \frac{0}{0} \end{align*} or \displaystyle \begin{align*} \frac{\infty}{\infty}  \end{align*}, then \displaystyle \begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \end{align*}.
    Well, that will save a lot of time.

    \lim x\rightarrow 0[\dfrac{e^{9x} - 1 - 9x}{x^{2}}]


    \lim x\rightarrow 0[\dfrac{e^{9(0)} - 1 - 9(0)}{(0)^{2}}]


    \lim x\rightarrow 0[\dfrac{e^{0} - 1 - 0}{0}]


    \lim x\rightarrow 0[\dfrac{1 - 1 - 0}{0} = \dfrac{0}{0}] - Indeterminate

    \lim x\rightarrow 0[\dfrac{e^{9x} - 1 - 9x}{x^{2}}] - looking at the original limit again

    \lim x\rightarrow 0[\dfrac{9e^{9x} -9}{2x}] - simple as pie Using L' Hospital's Rule

    \lim x\rightarrow 0[\dfrac{9e^{9(0)} -9}{2(0)}]

    \lim x\rightarrow 0[\dfrac{9 - 9}{0}] = \dfrac{0}{0} - Again indeterminate

    \lim x\rightarrow 0[\dfrac{9e^{9x} -9}{2x}] - Looking at first derivative again

    \lim x\rightarrow 0[\dfrac{9(9)e^{9x}}{2}]

    \lim x\rightarrow 0[\dfrac{(18e^{9x}}{2}]

    \lim x\rightarrow 0[\dfrac{(18e^{9(0)}}{2}] = 9 Look right as final answer? Computer says no.
    Last edited by Jason76; October 24th 2013 at 08:38 PM.
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    Re: L'Hospital's Rule Example with e

    Quote Originally Posted by Jason76 View Post
    \lim x\rightarrow 0[\dfrac{9(9)e^{9x}}{2}]

    \lim x\rightarrow 0[\dfrac{(18e^{9x}}{2}]
    What is 9 times 9 again?

    -Dan
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