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Math Help - Compute the Limit: Question 3

  1. #1
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    Compute the Limit: Question 3

    Is the following correct?


    \text{Calculate } \lim_{x \rightarrow -1 } \frac{\sqrt{x^2+8}-3}{x+1}

    = \lim_{x \rightarrow -1 } \frac{\sqrt{x^2+8}-3}{x+1 } \cdot \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}

    = \lim_{x \rightarrow -1 } \frac{x^2+8-9}{(x+1)(\sqrt{x^2+8}+3)}

    = \lim_{x \rightarrow -1 } \frac{x^2-1}{(x+1)(\sqrt{x^2+8}+3)}

    = \lim_{x \rightarrow -1 } \frac{(x-1)(x+1)}{(x+1)(\sqrt{x^2+8}+3)}

    = \lim_{x \rightarrow -1 } \frac{(x-1)}{(\sqrt{x^2+8}+3)}

    = \frac{(-1-1)}{(\sqrt{1^2+8}+3)}

    = -\frac{1}{3}
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  2. #2
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    Re: Compute the Limit: Question 3

    I did the same problem on my white board with out looking at your answer and I received the same thing.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Compute the Limit: Question 3

    Your answer is entirely correct!
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  4. #4
    Member sbhatnagar's Avatar
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    Re: Compute the Limit: Question 3

    Your answer is entirely correct but you could have done it easier and shorter.

    \lim_{x \to -1 }\frac{\sqrt{x^2+8}-3}{x+1}

    =\lim_{x \to -1 }\frac{2x}{2\sqrt{x^2+8}} (Using L' Hopital's Rule)

    =-\frac{1}{3}
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Compute the Limit: Question 3

    @sbhatnagar:
    You're right, but the problem is that for some reasons l'Hopital is used in cases where it's the only option and avoided in cases where you could solve the limit in a different way.
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  6. #6
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    Re: Compute the Limit: Question 3

    And, of course, here we do not know if Sparky has covered L'Hopital's rule or even differentiation yet in his class.
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