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Clarification on L'Hopital's Rule?

  1. Aug 1, 2012 #1
    The rule states that:

    [itex] { lim }_{ x\rightarrow c }\quad \frac { f(x) }{ g(x) } \quad =\quad { lim }_{ x\rightarrow c }\quad \frac { f'(x) }{ g'(x) } [/itex]

    Right?

    So if

    [itex] { lim }_{ x\rightarrow 2 }\frac { { x }^{ 2 }+1 }{ x-1 } \quad =\quad 5 [/itex]

    Then shouldn't

    [itex] { lim }_{ x\rightarrow 2 }\frac { { (x }^{ 2 }+1)' }{ (x-1)' } \quad =\quad 5 [/itex]

    Also equal to 5? However, it equals to 4. Can someone help me understand why?
     
  2. jcsd
  3. Aug 1, 2012 #2
    l'Hopital's rule only applies when the limit on the left-hand side would give 0/0 or some other indeterminate form when directly evaluated. If you try to use it otherwise, you get gibberish.

    Edit: Taylor expanding, on the other hand, will always give you a sensible result, and it makes l'Hopital's rule obsolete.
     
  4. Aug 1, 2012 #3
    What do you meal by taylor expanding? I think L'Hopital's rule is very far from obsolete.
     
  5. Aug 1, 2012 #4
    A Taylor series expansion writes a function in terms of its derivatives at a specific point.

    [tex]f(x) = f(a) + (x-a) f'(a) + \frac{1}{2!} (x-a)^2 f''(a) + \frac{1}{3!} (x-a)^3 f'''(a) + \ldots{}[/tex]

    This can be used to explain why l'Hopital's rule works (and when it doesn't). Say that we're taking the limit as [itex]x \to a[/itex] and that there are two functions such that [itex]f(a) = g(a) = 0[/itex]. Taylor expanding both functions gives us

    [tex]\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f(a) + (x-a) f'(a) + \ldots}{g(a) + (x-a)g'(a) + \ldots} = \lim_{x \to a} \frac{0 + (x-a)f'(a) + \ldots}{0 + (x-a)g'(a) + \ldots}[/tex]

    Provided that [itex]f'(a), g'(a) \neq 0[/itex] (or infinity, or something equally ludicrous), you can see how cancelling out [itex](x-a)[/itex] will give l'Hopital's rule. The [itex]\ldots[/itex] can be safely ignored provided that both first derivatives are nonzero, and if they are both zero, then you can look at the next non-vanishing terms instead.

    When [itex]f(a), g(a) \neq 0[/itex], you can read the limit off right away. Anything multiplied by [itex]x-a[/itex] will get very small compared to these constant terms as [itex]x \to a[/itex].

    Taylor series expansions are a powerful, surefire way to deal with limits, but in the end, given that you've not even heard of them yet, I wouldn't worry about trying to use them now. Just use l'Hopital's rule only when it's warranted (and not when it's improper to use), and you should be fine. When you get to Taylor series (I assume you will at some point), do consider them for evaluating limits (and lots of other useful things).
     
  6. Aug 1, 2012 #5
    I know what taylor expansions are, I just hadn't heard it used as a verb and I wasn't sure you were talking about that.

    I see where the taylor method comes from, but when you are dealing with more complex stuff and especially when you are dealing with calculating end point behavior/rates of change, I think L'Hopital's rule is much cleaner and nicer.
     
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