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Homework Help: Finding the limit? Piece function, stuck.

  1. Jan 21, 2008 #1

    I don't know what to do, and am stuck on this question.

    Let f(x) =
    {x^2+4, x does not equal 1
    {1, x=1 (right?)

    Lim F(x)

    First it gives a graph.
    A parabola but at x=1, y=5, there is an open dot, (hole in the graph) so it cannot equal 5?
    and then there is the random black dot at x=1, y=1.

    ok so I follow the graph from the left and the right and I get 5, but there is a hole there, and a random black dot on the graph not part of the parabola.

    as x approaches 1, y approaches...? (I want to say 5, but there's that darn hole!)

    But I am confused bc of the left part "x^2+4, x does not equal 1"

    What do I do when it "doesn't equal 1" but the limit says x->1?

    If it says lim x->1, yet CAN'T be 1 for the left side, then what do I do?!

    So can someone explain this problem? Is it just "does not exist"?
    Last edited: Jan 21, 2008
  2. jcsd
  3. Jan 21, 2008 #2


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    Science Advisor

    No, this limit very definitely exists and is simple.

    Do you remember the definition of "limit":

    [itex]\lim_{x\rightarrow a} f(x)= L[/itex] if and only if, for any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]0< |x-a|< \delta[/itex] then [itex]|f(x)- L|< \epsilon[/itex].

    In particular, look at that "0< |x- a|". The value of f at x= a, or even whether f(a) is defined is irrelevant!

    The function must be close to the limit for x close to but not equal to a. Now, suppose x is very very close to 1 but not equal to 1. What is f(x) close to?
  4. Jan 21, 2008 #3
    It looks like it is getting "closer" to 5.

    I learned that definition 1 week ago. These holes and graphs have made me so confused... esp in other sections we just learned, when you have the "removable discontinuity." We only use that if we have a fraction and it makes the denominator equal to zero (after we simplify), right?

    I'm confusing the sections in the book with each other.

    But for this problem, should I use lim x->1.001, and lim x->0.999? or the definition you gave me for limit(we've used that for proofs)?
    I'm just not sure when to use what method always.

    There's another problem like it asking for the limit, but it's rational, and the directions are the near the same for each, so now I'm a little confused.

    But I have only been in class 2 weeks, so I will keep reviewing and hope it will embed itself in my brain soon.
    Last edited: Jan 21, 2008
  5. Jan 21, 2008 #4


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    Science Advisor

    No, that happens any time you have a point at which the limit exists but is not equal to the value of the function there.

    For example: f(x)= 3x if x is not 1, 10000 if x= 1. Obviously (I hope it's obvious!) the limit of that function, as x goes to 1, is 3. Since f(1)= 10000, not 3, the function is not continuous there. But we can "remove" that discontinuity just by redefining f(1) to be 3. (Of course, then f(x) is the simple f(x)= 3x.)

    Why in the world would you use "lim x-> 1.001" and "lim x-> 0.999" when you were told to use "lim x-> 1"? My point was that you can ignore what happens at 1 but what happens arbitrarily close to 1 (1.0000000000000000000001 and even closer!) is important.

  6. Jan 21, 2008 #5
    Oh. I used it bc I was looking through my class notes, and that's what the professor did.
    But I guess he was showing us they both were getting closer to the same number (from the left and from the right...).
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