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Substituting functions in limits

  1. Dec 12, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm trying hard to understand as my professor hasn't taught(nor does my textbook) on how this works.

    It is known that $$\lim_{x \to 0}\frac{f(x)}{x} = -\frac12$$

    Solve
    $$\lim_{x \to 1}\frac{f(x^3-1)}{x-1}.$$

    2. Relevant equations


    3. The attempt at a solution
    OK.. so I do this:
    $$\begin{align}\lim_{x \to 1}\frac{f(x^3-1)}{x-1} &= \lim_{x \to 1}\frac{f(x^3-1)}{x-1}\cdot\frac{x^2+x+1}{x^2+x+1} \\ &= \lim_{x\to 1}\frac{f(x^3-1)}{x^3-1}(x^2+x+1) \\ &= \lim_{x\to 1}\frac{f(x^3-1)}{x^3-1}\lim_{x\to 1}(x^2+x+1) \\ &= \lim_{u\to 0}\frac{f(u)}{u}(3) \end{align}$$
    $$(3)(-1/2)$$

    Now my question is... why is it that $$lim_{u \to 0}\frac{f(u)}{u} = \lim_{x \to 0}\frac{f(x)}{x} = -\frac12 $$

    I thought that the limit changes when you substitute variables??? So when we substitute ##u = x^3 -1## the limit becomes u approaches 0 as when you substitute x=1 in to the above, you get u = 0. Am I wrong on this??

    Note that the answer is -3/2.
     
    Last edited: Dec 12, 2016
  2. jcsd
  3. Dec 12, 2016 #2

    fresh_42

    Staff: Mentor

    You may name the variable whatever you like:
    Whether it's ##\lim_{u \to 0}\frac{f(u)}{u}\;## or ##\; \lim_{\text{ tree} \to 0}\frac{f(\text{tree})}{\text{tree}}\;## or ##\;\lim_{x \to 0}\frac{f(x)}{x}##

    Or wasn't that the question and it's about the previous steps?
     
  4. Dec 12, 2016 #3
    I was asking why is it that lim u-> 0 f(u)/u = lim x -> 0 f(x)/x in the last step of the solution.
     
  5. Dec 12, 2016 #4

    fresh_42

    Staff: Mentor

    It is simply the same, identical. First, ##u## is simply an abbreviation of ##x^3-1##. But then we forget about the origin of ##u##, because it doesn't matter anymore. Once we have it, we can concentrate on the expression ##\lim_{u \to 0}\frac{f(u)}{u}## and we are free to rename it.
    If it feels better for you, then how about changing the given condition for ##f## to ##\lim_{v \to 0}\frac{f(v)}{v}=-\frac{1}{2}## and substitute another time ##v=u## at the end of the proof?
     
  6. Dec 12, 2016 #5
    If it is the same, then that's enough for me to know. Thanks!
     
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