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Spivak problem on limits

  1. Dec 29, 2015 #1
    Consider the limit
    lim f(x)g(x)
    x→a
    Spivak has proved that this is equal to lim f(x) multlied by
    x→a
    lim g(x)
    x→a

    And also if lim g(x) = k and k≠0,
    x→a
    Then. lim 1/g(x) = 1/k
    x→a

    Now the problem arises.....
    Consider the limit
    lim ((x^2)-(a^2))/(x-a)
    x→a
    It can factorised and written as( taking x-2 from numerator)
    lim (x+a)
    x→a
    Which is nothing but 2a.
    Now we can write it the above limit also as
    lim(x^2)-(a^2) multiplied by
    x→a
    lim 1/(x-a)
    x→a.

    The second limit does not exist because
    lim(x-a)=0 and l=0
    x→a
    So, its reciprocal limit does not exist.
    Then can't we say
    lim ((x^2)-(a^2))/(x-a) does not exist?
    x→a
    Where am I wrong in my arguement?
     
  2. jcsd
  3. Dec 29, 2015 #2

    Samy_A

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    ##\displaystyle\lim_{x\rightarrow a} {f(x)g(x)}=\lim_{x\rightarrow a} {f(x)}.\lim_{x\rightarrow a} {g(x)}## if the two limits in the RHS exist.
     
  4. Dec 29, 2015 #3
    Take ##f(x) = 1/x## and ##g(x) = 1/x^2##. By ##x\to\infty## both limits are zero. Then ##\lim_{x\to\infty}\frac{f(x)}{g(x)} =
    \lim_{x\to\infty}\frac{g(x)}{f(x)} ## even exist or not. Change now reference system to ##x\to{x-a}##. Now limits are ##-1/a## and ##1/a^2##. Can existance of ##\lim_{x\to\infty}\frac{f(x)}{g(x)}## depends by reference system?
     
  5. Dec 29, 2015 #4
    But, i have not started on 'limit tends to infinity part'. I will start soon.
     
  6. Dec 29, 2015 #5

    Samy_A

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    Limit to infinity or not is irrelevant. The product rule assumes that the two limits exist.

    Let's take a trivial example:
    ##f(x)=x-a##, ##g(x)=\frac{1}{x-a}##
    Then ##\displaystyle\lim_{x\rightarrow a} {f(x)g(x)}=\displaystyle\lim_{x\rightarrow a} {1}=1##.
    But ##\displaystyle\lim_{x\rightarrow a} g(x)= \lim_{x\rightarrow a} \frac{1}{x-a}## doesn't exist, so the expression ##\displaystyle\lim_{x\rightarrow a} f(x).\displaystyle\lim_{x\rightarrow a} g(x)## is not defined.
     
  7. Dec 29, 2015 #6
    Yes, i get it. The original condition is the limits should exist. He stated before proving it.
    A function f can be written as
    f=gb or cd...where g,b, c and d are different functions of x.
    let the limit of b as x approaches a, not exist. but for g,c,d the limts exist as x approaces a.
    How can we justify the fact that limit of f as f approaches a
    is nothing but (limit of c as x approaches a)×( limit of d as x approaches a)
    and not (limit of g as x approaches a)×(limit of b as x approaches a)?
     
  8. Dec 29, 2015 #7

    Samy_A

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    We justify it by noting that in your last expression, you "multiply" by something that doesn't exist.
     
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