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Math Help - L'Hospital's Rule Square Root Problem

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    L'Hospital's Rule Square Root Problem

    Find the limit as x approaches 1.

    Step 1)

    Plug in 1 into x.

     \frac{x - 1}{\sqrt{x^{2} + 3} - 2}

    This come out to being indeterminate or 0/0.

    So taking the rule:

    Step 2).  \frac{1}{\frac{x}{\sqrt{x^{2} + 3}}}

    Step 3) \frac{\sqrt{x^{2}+ 3}}{x} Final Answer is 2.

    Anyhow, I see the pattern of how the answer came about, but I don't understand the reasoning behind it. Well, I understand differintiation part. However. where did the fraction inside of the fraction come from in Step 2? Why did the square root in Step 2 get moved to the numerator in the final answer?
    Last edited by Jason76; November 9th 2012 at 02:50 PM.
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Jason76 View Post
    \frac{x - 1}{\sqrt{x^{2} + 3} - 2}
    This come out to being indeterminate.
    No it does not.
    \frac{x - 1}{\sqrt{x^{2} + 3} - 2}=\frac{(x - 1)(\sqrt{x^{2} + 3} +2)}{x^2-1}
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    Re: L'Hospital's Rule Square Root Problem

    @Jason76, what is the derivative of \sqrt{x^2 + 3} - 2 with respect to x?
    Last edited by richard1234; November 9th 2012 at 02:51 PM.
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Plato View Post
    No it does not.
    \frac{x - 1}{\sqrt{x^{2} + 3} - 2}=\frac{(x - 1)(\sqrt{x^{2} + 3} +2)}{x^2-1}
    When factored out it comes to 0 in the bottom. However, when not factored it's indeterminate. But either way, L'Hospital's rule is applied.
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Jason76 View Post
    When factored out it comes to 0 in the bottom. However, when not factored it's indeterminate. But either way, L'Hospital's rule is applied.
    No it does not. Please learn some basic algebra.
    \frac{(x - 1)(\sqrt{x^{2} + 3} +2)}{x^2-1}=\frac{\sqrt{x^{2} + 3} +2}{x+1}
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Plato View Post
    No it does not. Please learn some basic algebra.
    \frac{(x - 1)(\sqrt{x^{2} + 3} +2)}{x^2-1}=\frac{\sqrt{x^{2} + 3} +2}{x+1}
    In the book, the limit comes out to 2 (via using L' Hospital's rule). In your reasoning (via factoring), it's 3. Plugging 1 into the original equation (without factoring or using the rule), the answer is 0/0 or indeterminate. Something's wrong here.
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Jason76 View Post
    In the book, the limit comes out to 2 (via using L' Hospital's rule). In your reasoning (via factoring), it's 3. Plugging 1 into the original equation (without factoring or using the rule), the answer is 0/0 or indeterminate. Something's wrong here.
    How'd you get 3? Either way I got an answer of 2.

    As to your questions in your original post, they can both be answered by differentation, and basic algebra, which you should already know.
    Last edited by richard1234; November 9th 2012 at 04:39 PM.
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Jason76 View Post
    In the book, the limit comes out to 2 (via using L' Hospital's rule). In your reasoning (via factoring), it's 3. Plugging 1 into the original equation (without factoring or using the rule), the answer is 0/0 or indeterminate. Something's wrong here.
    As I said, Please learn to use basic algebra.
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    Re: L'Hospital's Rule Square Root Problem

     \frac{x - 1}{\sqrt{x^{2} + 3} - 2}


     \frac{x - 1}{\sqrt{x^{2} + 3} - 2} \frac{\sqrt{x^{2} + 3} + 2}{\sqrt{x^{2} + 3} + 2}

    \frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} + 3 - 4}

    \frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} - 1}

    \frac{(x - 1) (\sqrt{x^{2} + 3)} + 2)}{(x - 1) (x + 1)}

    \frac{\sqrt{x^{2} + 3} + 2)}{(x + 1)}

    Plugging in 1 gets 2 for the limit.
    Last edited by Jason76; November 9th 2012 at 05:11 PM.
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    Re: L'Hospital's Rule Square Root Problem

    What Plato's saying is, you probably messed up the arithmetic or something when you obtained an answer of 3 (did you forget the square root sign?). Also, what is wrong with the answer being indeterminate and 2? 0/0 is indeterminate because it could equal anything; that's why we use other techniques (L'Hopital's rule or plain algebra) to determine the limit, which is 2.

    Let's return back to the original solution, using L'Hopital's rule. Step 2 in your original solution should be easy if you know how to find derivatives. To get to step 3, you should also know that.
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by richard1234 View Post
    What Plato's saying is, you probably messed up the arithmetic or something when you obtained an answer of 3 (did you forget the square root sign?). Also, what is wrong with the answer being indeterminate and 2? 0/0 is indeterminate because it could equal anything; that's why we use other techniques (L'Hopital's rule or plain algebra) to determine the limit, which is 2.

    Let's return back to the original solution, using L'Hopital's rule. Step 2 in your original solution should be easy if you know how to find derivatives. To get to step 3, you should also know that.
    Answer is above your post.
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Jason76 View Post
    \frac{\sqrt{x^{2} + 3})}{(x + 1)}

    Plugging in 1 gets 2 for the limit.
    If you plug in 1 with the expression you provided, you get 1. Not 2.

    You're missing a "+2" in the numerator.
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    Re: L'Hospital's Rule Square Root Problem

    Went back and edited the post, it came out to 3, Plato was right. However, the book says 2. If differentiated using L Hospital's rule, then it comes out to 2. The book is from a major bookstore. I guess it's wrong.

    Edited Post:

     \frac{x - 1}{\sqrt{x^{2} + 3} - 2}


     \frac{x - 1}{\sqrt{x^{2} + 3} - 2} [\frac{\sqrt{x^{2} + 3} + 2}{\sqrt{x^{2} + 3} + 2}]

    \frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} + 3 - 4}

    \frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} - 1}

    \frac{(x - 1) (\sqrt{x^{2} + 3)} + 2}{(x - 1) (x + 1)}

    \frac{\sqrt{x^{2} + 3} + 2}{(x + 1)}

    Plugging in 1 gets 3 for the limit.
    Last edited by Jason76; November 9th 2012 at 05:24 PM.
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Jason76 View Post
    Went back and edited the post, it came out to 3, Plato was right. However, the book says 2. If differentiated using L Hospital's rule, then it comes out to 2. The book is from a major bookstore. I guess it's wrong.

    Edited Post:

     \frac{x - 1}{\sqrt{x^{2} + 3} - 2}


     \frac{x - 1}{\sqrt{x^{2} + 3} - 2} \frac{\sqrt{x^{2} + 3} + 2}{\sqrt{x^{2} + 3} + 2}

    \frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} + 3 - 4}

    \frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} - 1}

    \frac{(x - 1) (\sqrt{x^{2} + 3)} + 2)}{(x - 1) (x + 1)}

    \frac{\sqrt{x^{2} + 3} + 2)}{(x + 1)}

    Plugging in 1 gets 3 for the limit.
    As I said learn algebra. Now I say Learn basic arithmetic.
    \frac{\sqrt{1^{2} + 3} +2}{1+1}=2
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    Re: L'Hospital's Rule Square Root Problem

    Quote Originally Posted by Jason76 View Post

    \frac{\sqrt{x^{2} + 3} + 2}{(x + 1)}

    Plugging in 1 gets 3 for the limit.
    Nope.
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