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Introduction to limits problem

  1. May 16, 2008 #1
    in my calculus introduction book, it is written:

    Let a rational function: [tex]f(x) = \frac{{p(x)}}{{q(x)}}[/tex] and a, a real number

    If q(a) equals 0, but not p(a), then [tex]{\lim }\limits_{x \to a} f(x)[/tex] does not exist.

    however, while doing exercises on the internet, i found that:
    [tex]{\lim }\limits_{x \to 1} \frac{{2 - x}}{{(x - 1)^2 }} = \infty[/tex]

    is my textbook wrong?
    thank you
  2. jcsd
  3. May 16, 2008 #2
    Would you say that the other limit exist?
  4. May 16, 2008 #3


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    Saying that [itex]\lim_{\stack{x\rightarrow a}}f(x)= \infty[/itex] just means that the limit does not exist in a particular way.
  5. May 16, 2008 #4
    Let's convert what you have found on the net into the formula in your textbook. so q(x) = (x-1)^2 , p(x) = 2 - x and x--> 1. From here, we can say that q(1) = 0 and p(1) = 1. If you calculate p(1) / q(1), you get infinitity. It means that this limit does not exist.
  6. May 18, 2008 #5
    As x goes to 1, the denominator goes to zero (since, of course, (x-1)^2 = (1-1)^2 = 0. The smaller the denominator, the larger the output of the function. Since x makes the denominator shrink, we can rephrase the limit to be: "as denominator goes to zero, the output goes to a very large number."

    Lets look at a different, more simple function for a moment: f(x) = 1/x. As x becomes very small, the output ("y") becomes very very large. So what happens when x goes to exactly zero? This is hard to get out head around, so instead of saying "what happens when x is zero," we shall make an 'inverse function.' Now we will take OUTPUTS, and recieve x INPUTS. That is, large numbers will yield small numbers (where the small number, when put under a one, yields the original large number.)

    y= 1/x ; now use algebra to move the terms: x = 1/y.

    Now when y becomes large, x becomes small -- this is the inverse of the original function, now we START with the large number, y, and receive the small number x in return. What value of y will make x become zero? As y becomes increasingly large, x becomes increasingly small.

    We can now say:

    \lim_{\stack{y\rightarrow a}}f(y)= 0

    The question is, what value of a will make f(y) become zero? No matter HOW large we make y, x will still not be EXACTLY zero. It will become very close to zero, but it will not actually BE zero. Hence when x is actually 1 in your original example, we say that the output is "undefined." However, it is useful for practical purposes to say it is "infinity," since as x goes to 1, the output becomes increasingly large -- it's a way of saying the output becomes large without bound, even if the actual function is undefined at the value x is approaching (in this case, we get a divide by zero error.)

    In this sense, as y goes to a large number, x goes to a small number. We say the limit is "zero" or "infinity" because zero and infinity signify the apex of each extreme. There is a very ambiguous nature behind limits and infinity, so beat yourself up if you can't get your head around it (I don't think many people do, if any.)

    As a side note, there is a lot of inquiry as to the nature of infinity. You may be interested in reading about set theory and the extended complex plane -- they make some odd use of the notion of "infinity," though I'm unsure how to interpret it myself so I'll avoid going into detail here.
    Last edited: May 18, 2008
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