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Trying to understand limits.

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data
    lim(x+3)=?
    x→3

    2. Relevant equations
    Substituting x=3, I think.


    3. The attempt at a solution
    lim(x+3)= 3+3= 6
    x→3

    So we have got this straight lined graph, f(x)=x+3, and the answer I got shows... what exactly?
     
  2. jcsd
  3. Aug 1, 2012 #2

    Mark44

    Staff: Mentor

    Your answer is correct. The problem is extremely simple, so it doesn't really show much about the power of limits. What this is saying is that if x is some value close to 3, then x + 3 will be close to 6.

    Here's an example that is more informative.

    $$ \lim_{x \to 3} \frac{x^2 - 9}{x - 3}$$

    If you simply substitute 3 for x as you did in the problem you posted, you will get 0/0, which is not defined. Even so, the limit above actually exists, and is the same number as in your problem, 6.
     
  4. Aug 2, 2012 #3
    Exam problems are usually 0/0, ∞/∞, 1, 0, ∞0 which you can't just simply plug and chug.:devil:
     
  5. Aug 2, 2012 #4
    I agree, lim x->3 of x+3 really isn't very fun or useful...
    The power of limits lies in places, like Mark44 said, where you cannot simply substitute in some value.

    The limit is 'kind of' the answer you would expect to get if you nudge yourself infinitely close to a point.
    Take the function f(x) = x if x≠2 and x=0 if x=2.

    What is the limit of f(x) as x->2?
    Plugging in x=2 gives us f(2) which is 0. The limit is, however, 2.
    Another common example is limit as x->0 of Sin(x)/x, which happens to be 1 but it's not so obvious from just looking at it, and just plugging in x=0 gives you the lovely, undefined 0/0.

    The most common definition given for the limiting process is this;
    The limit of f(x) as x goes to a is L if;
    For any given ε>0 I can find a δ>0 such that whenever |x-a| < δ then |f(x) - L| < ε
     
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