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Math Help - Finding limit/derivative

  1. #1
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    Finding limit/derivative

    I have two questions, both are kind of similar involving rationalization and I would appreciate if someone can push me in the correct direction.

    The first is finding the limit of this:
    sqrt(3x-5) - 4
    x-7

    And this is as limit x-> 7

    I tried to rationalize and got this in the end

    3(x-7)
    (x-7)(sqrt(3x-5) - 4)

    But after canceling out the (x-7), the denominator is still 0. I'm not sure where I can go from there, I think I must have made a mistake somewhere..


    _______________________________________


    The second question is specifically finding the slope to the tangent at a given point from first principles.

    of y = sqrt{2x+8} at (-2,2)

    So this was the form I used,

    Limit h->0

    sqrt{2(2+h)+8} - 2
    h

    And then after trying to rationalize I get this

    2(2+h) + 8 - 4
    h(sqrt(2(2+h) +8) +2)

    I don't see where I can go from here either. The top doesn't leave something in kh form so I can't cancel it out....Did I make a mistake or approach it from the wrong direction?
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  2. #2
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    Quote Originally Posted by theowne View Post
    I have two questions, both are kind of similar involving rationalization and I would appreciate if someone can push me in the correct direction.

    The first is finding the limit of this:
    sqrt(3x-5) - 4
    x-7

    And this is as limit x-> 7

    I tried to rationalize and got this in the end

    3(x-7)
    (x-7)(sqrt(3x-5) - 4) Mr F says: That - should be a +.

    But after canceling out the (x-7), the denominator is still 0. I'm not sure where I can go from there, I think I must have made a mistake somewhere..


    _______________________________________


    The second question is specifically finding the slope to the tangent at a given point from first principles.

    of y = sqrt{2x+8} at (-2,2)

    So this was the form I used,

    Limit h->0

    sqrt{2(2+h)+8} - 2
    h

    Mr F says: x = -2 so the bit in red should be -2. So you need to re-work the solution from here.

    [snip]
    ..
    Last edited by mr fantastic; March 25th 2008 at 04:53 PM. Reason: Fixed a typo.
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  3. #3
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    I don't understand why it should be -(-2). The form is f(x+h) - f(x) correct (over h)? The components say (-2,2) so isn't f(x)=2?
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    Quote Originally Posted by theowne View Post
    I don't understand why it should be -(-2). The form is f(x+h) - f(x) correct (over h)? The components say (-2,2) so isn't f(x)=2?
    I made a typo. I've edited my reply. Take another look now.

    x = -2 so you need f(-2 + h) - f(-2). You have f(2 + h) - f(-2).
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    Thanks for the help. I seem to make these kind of simple errors often.
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  6. #6
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    If I may ask one more question, I have one more example:

    2x^3 - 4x
    3x^4 + 2x^3

    I'm not too sure how to approach this question. I'm guessing it involves factoring but I'm not sure exactly what..
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  7. #7
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    Quote Originally Posted by theowne View Post
    If I may ask one more question, I have one more example:

    2x^3 - 4x
    3x^4 + 2x^3

    I'm not too sure how to approach this question. I'm guessing it involves factoring but I'm not sure exactly what..
    New questions belong in new threads!

    I presume you need to simplify this?
    \frac{2x^3 - 4x}{3x^4 + 2x^3}

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

    Note that the numerator can be factored further, but as nothing more will cancel I didn't worry about doing it.

    -Dan
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  8. #8
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    Hi,

    Need to find the limit as x approaches 0. Sorry, I didn't realize I'd need a new thread since this question was similar to the originals. Simplified that way, the denominator would still be zero, wouldn't it?
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by theowne View Post
    Hi,

    Need to find the limit as x approaches 0. Sorry, I didn't realize I'd need a new thread since this question was similar to the originals. Simplified that way, the denominator would still be zero, wouldn't it?
    Yes, and as the limit in the numerator does not go to zero, then the limit does not exist.

    -Dan
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  10. #10
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    Quote Originally Posted by topsquark View Post
    New questions belong in new threads!

    I presume you need to simplify this?

    [snip]
    -Dan
    And ideally you will post the whole question so that presumptions/assumptions are unnecessary.
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