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Limit question

  1. Feb 23, 2009 #1
    so i understand how to resolve a limit at x->oo, but from a conceptual standpoint, i do not get it. for example,
    limit x->oo, 4x/5x
    so the answer is 4/5, but oo/oo is an indeterminate expression
    i understand that if i treat x as a variable, then it makes sense, but still
    if the example was 4x/5x and i said that x was infinity, then there would be no answer.
    in addition
    4*oo = 5*oo = oo
    so following that method, it would return one as the answer every time there were the same exponents on the x (so in other words, a*x/(b*x)

    can someone please explain this to me?
    thanks.
     
  2. jcsd
  3. Feb 23, 2009 #2
    For every non-zero value of x, 4x/(5x) = 4/5.
    An indeterminate expression is one for which further analysis of the components is necessary. It does not imply that the limit does not exist.
     
  4. Feb 23, 2009 #3
    Since [itex] \frac{\infty}{\infty}[/itex] is indeterminate, you can use L'Hopital's rule which says that [itex]\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}[/itex] iff [itex]\lim_{x \to a} \frac{f(x)}{g(x)}[/itex] is interterminate in the form [itex] \frac{0}{0}[/itex] or [itex]\frac{\infty}{\infty}[/itex] and [itex]\lim_{x \to a} \frac{f'(x)}{g'(x)}[/itex] exists.


    [itex] \lim_{x \to \infty} \frac{4x}{5x} = \lim_{x \to \infty} \frac{(4x)'}{(5x)'} = \lim_{x \to \infty} \frac{4}{5} = \frac{4}{5}[/itex]

    Of course, that's not necessary because you can simply cancel the x's, but it's useful if you run into things like these.
     
  5. Feb 23, 2009 #4
    can you prove that?
     
  6. Feb 24, 2009 #5

    HallsofIvy

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    Prove what, L'Hopital's rule? Tedious, but the proof is found in any Calculus text book.

    For this simple problem, it is better to note that in the definition of "limit at infinity",
    "[itex]\lim_{x\rightarrow \infty} f(x)= L[/itex] if and only if, for any [itex]\epsilon> 0[/itex], there exist N such that whenever x > N, |f(x)- L|<[itex]\epsilon[/itex]"
    x is always a real number- you never "set x= infinity". For any real number x, other than 0, 4x/5x= 4/5 so |4x/5x- 4/5|= |4/5- 4/5|= 0 which is less than any positive [itex]\epsilon[/itex].

    It is also interesting to note that, although setting x= 0 gives an "indeterminant", the limit there is also 4/5. For finite a, the definition of limit is
    "[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(a)- L|< \epsilon[/itex]."

    Notice that "0< |x- a|". What happens AT x= a is irrelevant. for all x such that 0< x, 4x/5x= 4/5 so 4/5 is the limit.
     
  7. Feb 24, 2009 #6

    CRGreathouse

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    That's why the question is "what is the limit of the expression as x approaches infinity", not "what is the value of this expression evaluated at x = infinity".
     
  8. Feb 24, 2009 #7
    related, but is limit x->oo a finite number?
     
  9. Feb 24, 2009 #8

    HallsofIvy

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    Limit of what function? It has already been made clear that the limit, as x goes to infinity of the particular expression here, 4x/5x, is 4/5. If it were instead (4x+ 5)/(5x- 2) the limit would still be 4/5. If it were 4/5x, the limit would be 0. If it were 4x/5, then there would be no limit.

    (Strictly speaking, because the limit must be a real number, ANY limit is a finite number. We should say "the limit does not exist" rather than "the limit is infinity". (Some textbooks use the phrase "diverges to infinity".)
     
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