Is there a website or can someone who help me with finding a limit-calculus? Thanks?

Update:

The derivative-

Example: find lim (x arrow pointing at 2 under lim) g(x), where g(x)=2^2+4/x-2

Update 2:

Sorry I did write it wrong. It is suppose to be:

Find lim x-->2 g(x), where g(x)=(x^2+4)/(x-2).

Update 3:

I did it wrong again. I don't know what is wrong with me. I am looking at the problem and typing something different. One more time:

Find lim x-->2 g(x), where g(x)=(x^2+4)/(x-2)

4 Answers

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  • 1 decade ago
    Favorite Answer

    try this websites:

    www.mathnstuff.com

    www.calculus-help.com (watch)

    www.algebralab.org

    www.stetson.edu

    just try it

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  • 1 decade ago

    http://www.math.com/homeworkhelp/Calculus.html

    http://amby.com/educate/math

    http://www.webmath.com

    Um... i just saw your example, are you missing some variables and/or parenthesis?

    The lim with the x arrow pointing at 2 is ok, I get your meaing there. But the way you have it written, do you mean 2^2 + (4/x) -2?

    In that case, simply substitute the 2 and you get 4 + (4/2) -2 = 4 + 2 -2 = 4

    Or do you mean 2^2 + 4/(x-2)?

    In that case you get 4 + 4/0, which is infinite or undefined. (graph it and you'll see a vertical asymptote at 2, with the graph going to positive infinity on the right side of the asymptote and negative infinity on the left side.)

    Or is there supposed to be an x with the 2, did you mean (2x^2 + 4)/(x-2)?

    In that case, you still have a vertical asymptote with the function going to negative infintiy on the left side of 2 and positive infinity on the right, but the effect is more pronounced because of the square in the numerator.

    When in doubt, use parenthesis.

    Ok, I just saw your editing, and that still has a vertical asymptote at 2 and behaves similar to the other functions discussed here. So the answer is still limit undefined.

    Generally, what you do with these is, you try and factor the numerator and denominator and cancel out common factors. If you can do this, you have what is called a "removable discontinuity" and so you can find a limit. Otherwise you have a "non removable discontinuity", which shows up as a vertical asymptote on the graph, and there's no limit.

    Now if you'd had x^2 - 4 in the numerator you would have been able to remove the discontinuity. but with x^2 + 4 in the numerator, you're stuck with it.

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  • 1 decade ago

    lim (x --> 2) (2*2 + 4/(x-2))

    In this case is a simple limit: you just have to think what happens if x is close to 2, and continue reasoning based on it.

    First of all, if x --> 2, then (x-2) --> 0

    so, 4/(x-2) --> 4/0, e.g. infinite

    if x > 2, then (x-2) is positive and so the limit is +infinite

    if x < 2, then (x-2) is negative and so the limit is -infinite

    the extra 2*2 added to +infinite or -infinite has no effect, so we are finished.

    The conclusion is:

    if x --> 2 "by the left" (e.g. with x < 2), then the limit is -infinite

    if x --> 2 "by the right" (e.g. with x > 2), then the limit is +infinite

    if, in general, we just know that x --> 2, then we cannot give any limite (because it is not clear if is +infinite or -infinite), and so the conclusion is that the limite DOES NOT EXIST.

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  • Philo
    Lv 7
    1 decade ago

    If you'd said what limit, I could've helped more.

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