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Math Help - Calculate the limit (Question 2)

  1. #1
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    Calculate the limit (Question 2)

    I'm stuck on the last line and I'm not sure how to finish:

    \text{Calculate }\lim_{x \rightarrow-2 } \frac{x+2}{\sqrt(x^+5) -3}

    = \lim_{x \rightarrow-2 } \frac{x+2}{\sqrt(x^+5) -3 } \cdot \frac{\sqrt (x^2-5) + 3}{\sqrt (x^2-5) + 3}

    = \left (\sqrt (x^2 +5) -3 \right ) \cdot \left (\sqrt (x^2 -5) +3 \right )

    = x^2 + 5 - 9

    = x^2 - 4

    = (x+2)(x-2)

    = \lim_{x \rightarrow-2 } \frac{(x+2) \cdot \sqrt (x^2 -5) + 3}{(x+2)(x-2)}

    = \lim_{x \rightarrow-2 } \frac{\sqrt (x^2-5)+3}{x-2}

    = \frac{\sqrt ((-2)^2-5)+3}{(-2)-2}

    = \frac{\sqrt (-1)+3}{-4 }Stuck here, adding square root of -1+3
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  2. #2
    Grand Panjandrum
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    Re: Calculate the limit (Question 2)

    Quote Originally Posted by sparky View Post
    I'm stuck on the last line and I'm not sure how to finish:

    \text{Calculate }\lim_{x \rightarrow-2 } \frac{x+2}{\sqrt(x^+5) -3}
    For this you just need to substitute -2 for x to get the limit is 0, unless everything in the denominator is under the square root

    CB
    Last edited by CaptainBlack; September 14th 2011 at 12:41 PM.
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    Re: Calculate the limit (Question 2)

    \text{Calculate } \lim_{x\rightarrow-2} \frac{x+2}{\sqrt(x^2 + 5)-3}

    = \frac{-2+2}{\sqrt(4^2 + 5)-3}

    =\frac{0}{3-3}

    = 0

    CaptainBlack, I have three questions:

    1. What does 0 really mean?
    2. What is the importance of plugging in -2 first?
    3. What if the answer was not 0? than what do I do?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Calculate the limit (Question 2)

    The importance of plugging in -2 first is that you can determine if you're dealing with an undetermined form or not, in this case you see you get the undetermined form: \frac{0}{0}.

    Calculating:
    \lim_{x \to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{(\sqrt{x^2+5}-3)\cdot (\sqrt{x^2+5}+3)}
    =\lim_{x \to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{x^2-4}
    =\lim_{x\to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{(x-2)\cdot (x+2)}
    =\lim_{x\to -2} \frac{\sqrt{x^2+5}+3}{x-2}

    Now plug in -2 again, so what's the limit?
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    Re: Calculate the limit (Question 2)

    Quote Originally Posted by Siron View Post
    The importance of plugging in -2 first is that you can determine if you're dealing with an undetermined form or not, in this case you see you get the undetermined form: \frac{0}{0}.

    Calculating:
    \lim_{x \to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{(\sqrt{x^2+5}-3)\cdot (\sqrt{x^2+5}+3)}
    =\lim_{x \to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{x^2-4}
    =\lim_{x\to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{(x-2)\cdot (x+2)}
    =\lim_{x\to -2} \frac{\sqrt{x^2+5}+3}{x-2}

    Now plug in -2 again, so what's the limit?
    Thanks Siron for pointing out where I went wrong in my working. Here is what I got:

    =\frac{\sqrt {-2^2 + 5} + 3}{-2-2}

    =\frac{\sqrt {9} + 3}{-4}

    =\frac{6}{-4}

    = - \frac{3}{2}

    I want to be clear on this and correct me if I am wrong: if I get an undetermined value when I plug in the value (e.g. -2), then I have to calculate the limit the long way, which is what we just did. Otherwise, if I don't get an undertermined value, then the answer is the limit?
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  6. #6
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    Re: Calculate the limit (Question 2)

    Quote Originally Posted by sparky View Post
    \text{Calculate } \lim_{x\rightarrow-2} \frac{x+2}{\sqrt(x^2 + 5)-3}

    = \frac{-2+2}{\sqrt(4^2 + 5)-3}

    =\frac{0}{3-3}

    = 0

    CaptainBlack, I have three questions:

    1. What does 0 really mean?
    2. What is the importance of plugging in -2 first?
    3. What if the answer was not 0? than what do I do?
    Where did the x^2 suddenly appear from? If what you now post is what you really want then plugging -2 in for x won't work because what you get is indeterminate.

    You will either have to use L'Hopital's rule or the substitution u=x+2 and consider the resulting limit as u goes to zero

    CB
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    Re: Calculate the limit (Question 2)

    Sorry Captain Black, I made a mistake with the question. The correct question is:

    \text{Calculate } \lim_{x \rightarrow -2} \frac{x+2}{ \sqrt{x^2 + 5}-3}

    \lim_{x \rightarrow -2} \frac{x+2}{ \sqrt{x^2 + 5}-3 } \cdot \frac{\sqrt{x^2 + 5}+3}{\sqrt{x^2 + 5}+3}

    (\sqrt{x^2+5} -3) \cdot (\sqrt{x^2+5}+3)

    x^2 +5 -9

    x^2 - 4

    (x+2)(x-2)

    \lim_{x \rightarrow -2 } \frac{(x+2) \cdot (\sqrt{x^2 + 5}+3)}{(x+2) \cdot (x-2)}

    \lim_{x \rightarrow -2 } \frac{\sqrt{x^2+5}+3}{x-2}

    = \frac{\sqrt{-2^2+5}+3}{-2-2}

    \frac{6}{-4}

    - \frac{3}{2}

    How is my working and answer?
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: Calculate the limit (Question 2)

    This working and answer is fine!
    Note: make sure you use brackets where necessary, because (-2)^2=4 but -2^2=-4.
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