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Math Help - Help with limits. I can do the limit, but I can't explain "which theorem i used"

  1. #1
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    Help with limits. I can do the limit, but I can't explain "which theorem i used"

    1. lim as x goes to 2...(x^2 - 4)/(x-2)

    2. lim as x goes to 0...sin 2x + x^2 cos 5x

    3. lim as x goes to 0 from the negative side... [(square root)x^2]/x


    as ive said, ive gotten all the limits, but I need to tell which theorem i used.... we can only use like the 5 most basic ones, and i cant see how any of them apply to any of these
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    MHF Contributor chisigma's Avatar
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    Identities like these...

     \frac{x}{x} = 1

    \sqrt{x^2}=x

    ... are not consequences of some particular theorem but only a question of elementary 'good sense'...

    Kind regards

    \chi \sigma
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    thats exactly what i thought, but they explicitly say to note what limit theorems you are using

    the ones we got are:
    basically the 4 operations using limits f and g...eg lim f(x) + g(x)= lim f(x) + lim g(X)

    squeeze principal

    continuity of indefinite inegrals
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    Quote Originally Posted by twostep08 View Post
    thats exactly what i thought, but they explicitly say to note what limit theorems you are using

    the ones we got are:
    basically the 4 operations using limits f and g...eg lim f(x) + g(x)= lim f(x) + lim g(X)

    squeeze principal

    continuity of indefinite inegrals

    (x^2 - 4)/(x-2) = (x-2)(x+2)/(x-2) . Now, from the very definition of limit, when x --> 2 we explicitly rule out the value x = 2, so that above we can divide by x-2 and get very simply lim (x + 2) = 4 when x --> 2

    For the enxt one you use arithmetic of limits: both sin(2x) and x^2*cos 5x have limit zero when x --> 0: the first one directly by continuity of sin x. and the second one because x^2 --> 0 clearly and cos 5x is bounded.

    The last one is nice: since we're working with negative x's, Sqrt(x^2) = |x| ==> the limit wanted is lim |x|/x = lim (-1) = -1 when x --> 0-

    Tonio
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    Quote Originally Posted by chisigma View Post
    Identities like these...

     \frac{x}{x} = 1

    \sqrt{x^2}=x

    ... are not consequences of some particular theorem but only a question of elementary 'good sense'...

    Kind regards

    \chi \sigma
    Neither of those are true.
    \frac{x}{x}= 1 only for x\ne 0 and \sqrt{x^2}= |x|, not x.
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    Quote Originally Posted by twostep08 View Post
    1. lim as x goes to 2...(x^2 - 4)/(x-2)
    Theorem: if f(x)= g(x) for all x except x= a, then \lim_{x\rightarrow a} f(x)= \lim_{x\rightarrow a} g(x).

    2. lim as x goes to 0...sin 2x + x^2 cos 5x
    Theorem: if f is continuous at x= a, then lim_{x\rightarrow a}f(x)= f(a)
    Well, more the definition of "continuous" than a theorem. And sine and cosine are continuous for all x.

    3. lim as x goes to 0 from the negative side... [(square root)x^2]/x
    \sqrt{x^2}= |x| which is -x for x negative. That is, \frac{\sqrt{x^2}}{x}= \frac{-x}{x} which is -1 as long as x is not 0. Now use the theorem I cited for (1).


    as ive said, ive gotten all the limits, but I need to tell which theorem i used.... we can only use like the 5 most basic ones, and i cant see how any of them apply to any of these
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    Quote Originally Posted by HallsofIvy View Post
    \sqrt{x^2}= |x| which is -x for x negative.
    This cannot be true, can it? Aren't you are saying |x|=-x when x is negative? This flies in the face of absolute values!
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    No it's true like it is. If x<0 then x^2>0 and so does the absolute value...
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    Quote Originally Posted by pflo View Post
    This cannot be true, can it? Aren't you are saying |x|=-x when x is negative? This flies in the face of absolute values!

    Not at all: this is the exact definition of abs. value for negative values...pay attention: NEGATIVE values , so -x is.....

    Tonio
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Neither of those are true.
    \frac{x}{x}= 1 only for x\ne 0 and \sqrt{x^2}= |x|, not x.
    I strongly agree on the principle that everibody is free to express his own opinion, no matter which opinion is...

    ... but I also strongly prefer avoid to wasting time in tedious discussions because...




    Kind regards

    \chi \sigma
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  11. #11
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    Quote Originally Posted by chisigma View Post
    I strongly agree on the principle that everibody is free to express his own opinion, no matter which opinion is...

    ... but I also strongly prefer avoid to wasting time in tedious discussions because...




    Kind regards

    \chi \sigma
    The post by HoI (which you quoted) is not an opinion, it is mathematical fact. Details like this are important. It is never a waste of time to note and correct errors of understanding. These sorts of errors are extremely common - it is not a waste of time to make sure that the OP has the correct mathematical understanding. Far from being tedious, the discussion in this thread is extremely valuable.


    Edit: Thread closed. The discussion was getting off-point. As far as I can see, the questions have been satisfactorily answered. twostep08, if you require more help please pm me.
    Last edited by mr fantastic; October 9th 2009 at 10:14 PM.
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