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Math Help - L'Hospital's Rule: Two Problems?

  1. #1
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    L'Hospital's Rule: Two Problems?

    I've been working on these two problems for some time and cannot get either of them to work. ( I don't know how to use Math HTML so bear with me )

    Find the Limit

    Problem 1: lim [ (x/x-1) - (1/lnx) ]
    x->1

    Problem 2: lim [ sqrt(x^2 + x) - x ]
    x->inf


    I've been working for quite a while. The solution to both problems is 1/2
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  2. #2
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    The X-> represents as x goes to
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    Quote Originally Posted by DocileRaptor View Post
    I've been working on these two problems for some time and cannot get either of them to work. ( I don't know how to use Math HTML so bear with me )

    Find the Limit

    Problem 1: lim [ (x/x-1) - (1/lnx) ]
    x->1

    Problem 2: lim [ sqrt(x^2 + x) - x ]
    x->inf


    I've been working for quite a while. The solution to both problems is 1/2
    Dear DocileRaptor,

    For the first problem,

    \lim_{x\rightarrow{1}}\left(\frac{x}{x-1}-\frac{1}{\ln{x}}\right)

    \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)

    When x=1 the numerator and denominator becomes zero. Hence you can use the L'Hospitals rule. Hope you can continue from here.

    For the second problem, multiply both the numerator and denominator by, \sqrt{x^2+x}+x. Then you would be able to use L'Hospital's rule.

    Hope this will help you.
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  4. #4
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    Red face

    Hello Sudharaka,

    I suppose I should have posted how far I did get. I got to this step.
    However, when I tried using L'Hospital's Rule, I end up with 0/2 which is incorrect. Can you show me a step process of solving this problem?

    \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)<br />

    I'll try that for the second one, multiplying by the reciprocal makes sense.
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  5. #5
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    For the second one, I end up with...

    <br />
\lim_{x\rightarrow{\infty}}\left(\frac{x}{\sqrt{x^  2+x}+x}\right)



    Not sure how this is equal to 1/2
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    Quote Originally Posted by DocileRaptor View Post
    Hello Sudharaka,

    I suppose I should have posted how far I did get. I got to this step.
    However, when I tried using L'Hospital's Rule, I end up with 0/2 which is incorrect. Can you show me a step process of solving this problem?

    \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)<br />

    I'll try that for the second one, multiplying by the reciprocal makes sense.
    Dear DocileRaptor,

    By L'Hospital's rule,

    \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)<br />

    =\lim_{x\rightarrow{1}}\left(\frac{1+\ln{x}-1}{\frac{x-1}{x}+\ln{x}}\right)

    When x=1, the numerator and denominator both becomes to zero. Hence you have to use the L'Hospital's rule again. Hope you can continue from here.
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    Quote Originally Posted by DocileRaptor View Post
    For the second one, I end up with...

    <br />
\lim_{x\rightarrow{\infty}}\left(\frac{x}{\sqrt{x^  2+x}+x}\right)



    Not sure how this is equal to 1/2
    Dear DocileRaptor,

    You are correct. The limit becomes,

    \lim_{x\rightarrow{\infty}}\frac{x}{\sqrt{x^2+x}+x  }

    Now you can use the L'Hospital's rule(since when x goes to infinity both the numerator and denominator goes to infinity).

    But there's an easy method. Divide both the numerator and denominator by x.....

    Hope this will help you.
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  8. #8
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    Quote Originally Posted by Sudharaka View Post
    Dear DocileRaptor,

    By L'Hospital's rule,

    \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)<br />

    =\lim_{x\rightarrow{1}}\left(\frac{1+\ln{x}-1}{\frac{x-1}{x}+\ln{x}}\right)

    When x=1, the numerator and denominator both becomes to zero. Hence you have to use the L'Hospital's rule again. Hope you can continue from here.
    Ah ha! I see now. This is the first day with L'Hospital's Rule.

    So using L'Hospital's Rule twice consecutively on

    \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)<br />

    I end up with

    (1/x) / (1+ (1/x) ) as X approaches 1 which is equal to 1/2 Can you put that equation into the code so I can see how the code looks? I am learning how to use latex as we speak
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    Quote Originally Posted by DocileRaptor View Post
    Ah ha! I see now. This is the first day with L'Hospital's Rule.

    So using L'Hospital's Rule twice consecutively on

    \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)<br />

    I end up with

    (1/x) / (1+ (1/x) ) as X approaches 1 which is equal to 1/2 Can you put that equation into the code so I can see how the code looks? I am learning how to use latex as we speak
    Dear DocileRaptor,

    \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)<br />
=\lim_{x\rightarrow{1}}\left(\frac{\frac{1}{x}}{1+  \frac{1}{x}}\right)=\frac{1}{2}
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  10. #10
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    Quote Originally Posted by Sudharaka View Post
    Dear DocileRaptor,

    You are correct. The limit becomes,

    \lim_{x\rightarrow{\infty}}\frac{x}{\sqrt{x^2+x}+x  }

    Now you can use the L'Hospital's rule(since when x goes to infinity both the numerator and denominator goes to infinity).

    But there's an easy method. Divide both the numerator and denominator by x.....

    Hope this will help you.
    Can you show me what happens when you divide by X?
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  11. #11
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    Quote Originally Posted by DocileRaptor View Post
    Can you show me what happens when you divide by X?
    Surely, you can do that. Show me your work and I shall help you.
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  12. #12
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    The top divided by x is x/x = 1
    the bottom is (sqrt(x^2 + x) / x ) + (x/x)

    So it becomes 1/(sqrt(x^2 + x) / x ) +1
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  13. #13
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    Quote Originally Posted by DocileRaptor View Post
    The top divided by x is x/x = 1
    the bottom is (sqrt(x^2 + x) / x ) + (x/x)

    So it becomes 1/(sqrt(x^2 + x) / x ) +1
    Dear DocileRaptor,

    If you simplify more,

    \frac{1}{\frac{\sqrt{x^2+x}}{x}+1}=\frac{1}{\frac{  \sqrt{x^2+x}}{\sqrt{x^2}}+1}=\frac{1}{\sqrt{1+\fra  c{1}{x}}+1}

    Hope you would be able to continue from here.
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    Doesn't the bottom become sqrt(2) + 1 ?

    I feel like I'm missing something...
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  15. #15
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    Quote Originally Posted by DocileRaptor View Post
    Doesn't the bottom become sqrt(2) + 1 ?

    I feel like I'm missing something...
    Dear DocileRaptor,

    \lim_{x\rightarrow{\infty}}\frac{1}{\sqrt{1+\frac{  1}{x}}+1}

    =\frac{1}{\sqrt{1+0}+1}=\frac{1}{1+1}=\frac{1}{2}
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