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Thread: 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...

    $\displaystyle \frac{x}{x} = 1$

    $\displaystyle \sqrt{x^2}=x$

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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...

    $\displaystyle \frac{x}{x} = 1$

    $\displaystyle \sqrt{x^2}=x$

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Neither of those are true.
    $\displaystyle \frac{x}{x}= 1$ only for $\displaystyle x\ne 0$ and $\displaystyle \sqrt{x^2}= |x|$, 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 $\displaystyle \lim_{x\rightarrow a} f(x)= \lim_{x\rightarrow a} g(x)$.

    2. lim as x goes to 0...sin 2x + x^2 cos 5x
    Theorem: if f is continuous at x= a, then $\displaystyle lim_{x\rightarrow a}f(x)= f(a)$
    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
    $\displaystyle \sqrt{x^2}= |x|$ which is -x for x negative. That is, $\displaystyle \frac{\sqrt{x^2}}{x}= \frac{-x}{x}$ 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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    Member pflo's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    $\displaystyle \sqrt{x^2}= |x|$ which is -x for x negative.
    This cannot be true, can it? Aren't you are saying $\displaystyle |x|=-x$ when x is negative? This flies in the face of absolute values!
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    No it's true like it is. If $\displaystyle x<0$ then $\displaystyle x^2>0$ 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 $\displaystyle |x|=-x$ 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.
    $\displaystyle \frac{x}{x}= 1$ only for $\displaystyle x\ne 0$ and $\displaystyle \sqrt{x^2}= |x|$, 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

    $\displaystyle \chi$ $\displaystyle \sigma$
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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

    $\displaystyle \chi$ $\displaystyle \sigma$
    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; Oct 9th 2009 at 09:14 PM.
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