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Proof of special case of l'Hôpital's rule

AP.CALC:
LIM‑4 (EU)
,
LIM‑4.A (LO)
,
LIM‑4.A.1 (EK)
,
LIM‑4.A.2 (EK)
L'Hôpital's rule helps us find limits in the form limit, start subscript, x, \to, c, end subscript, start fraction, u, left parenthesis, x, right parenthesis, divided by, v, left parenthesis, x, right parenthesis, end fraction where direct substitution ends in the indeterminate forms start fraction, 0, divided by, 0, end fraction or start fraction, infinity, divided by, infinity, end fraction.
The rule essentially says that if the limit limit, start subscript, x, \to, c, end subscript, start fraction, u, prime, left parenthesis, x, right parenthesis, divided by, v, prime, left parenthesis, x, right parenthesis, end fraction exists, then the two limits are equal:
limit, start subscript, x, \to, c, end subscript, start fraction, u, left parenthesis, x, right parenthesis, divided by, v, left parenthesis, x, right parenthesis, end fraction, equals, limit, start subscript, x, \to, c, end subscript, start fraction, u, prime, left parenthesis, x, right parenthesis, divided by, v, prime, left parenthesis, x, right parenthesis, end fraction
The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.
Khan Academy video wrapper
Proof of special case of l'Hôpital's ruleSee video transcript

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  • blobby green style avatar for user Gabriela Aslı Rino Nesin
    In the video of the proof: don't we need g'(a) to be nonzero in order to divide by it?
    (8 votes)
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    • starky sapling style avatar for user 20leunge
      Sorry about the confusion earlier but here's my new answer: Yes, it has to be nonzero in order to divide it. If g'(a) is zero AND f'(a) is zero, then it is indeterminate form and you can just use L'Hopital's Rule again. If f'(a) is not zero, then the limit probably doesn't exist. I hope this answers your question.
      (26 votes)
  • aqualine ultimate style avatar for user xyzPoKeFaNxyz
    Can I proof it when f(x) is infinity?
    (7 votes)
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    • blobby green style avatar for user Yaaan
      still let f(a)=0, g(a)=0. but m(x) =1/g(x), n(x) = 1/f(x) So that m(a)= ∞, n(a) = ∞.
      Then m'(a)/n'(a) =
      lim(x==>a) {(m(x)-m(a))/(x-a)} / {(n(x)-n(a))/(x-a)} =
      lim(x==>a) {m(x)-m(a)} / {n(x)-n(a)} =
      lim(x==>a) {1/g(x)-1/g(a)} / {1/f(x)-1/f(a)} =
      lim(x==>a) {[g(a)-g(x)]/[(g(x)g(a)]} / {[f(a)-f(x)]/[(f(x)f(a)]}
      g(a)= 0, f(a) = 0 in numerator, so We can symplify it, then
      lim(x==>a) {[g(a)-g(x)]/[(g(x)g(a)]} / {[f(a)-f(x)]/[(f(x)f(a)]} =
      lim(x==>a) {[-g(x)]/[(g(x)g(a)]} / {[-f(x)]/[(f(x)f(a)]} =
      f(a)/g(a) = m(a)/n(a)
      then we proved m'(a)/n'(a)=m(a)/n(a),
      then lim(x==>a) m(x)/n(x) = m'(a)/n'(a) when m(a), n(a) are all infinite
      (3 votes)
  • piceratops ultimate style avatar for user abhi.devata
    In the video, Sal works backwards from the formula that we use: lim(x==>c) of f(x)/g(x)=0/0 = lim(x==>c) of f'(x)/g'(x). But he starts the proof with the f'(x) and g'(x). That would mean that trial and error was performed. How do we prove this by starting with f(x) and g(x) instead of their derivatives?
    (3 votes)
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    • blobby green style avatar for user majumderzain
      Here's an example with less calculus:
      We want to prove that 4 = 16/8-7+9. Normally you would do:
      16/8-7+9 = 2-7+9 = -5+9 = 4. But you could also do this backwards:
      4 = -5+9 = 2-7+9 = 16/8-7+9.

      These proofs do the same thing, but one is more straightforward than the other. Sal chooses to start with lim(f'(x)/g'(x)) and get to lim(f(x)/g(x)) because the proof is more straightforward that way.
      (10 votes)
  • leafers tree style avatar for user R Haq
    Can't this qualify as a general proof? Since l'hopitals rule only qualifies for indeterminate forms (0/0 and infinity/ infinity) if you've proven it for 0/0 you've infinity as well.
    All indeterminate forms can be written as 0/0.
    For example, consider a limit
    lim_x->c f(x)/g(x)
    which is infinity over infinity, one can equivalently say
    lim_x->10^+ (1/g(x))/(1/f(x))
    which is 0/0
    and we've proven l'hopitals rule for 0/0
    you can see more convertions between 0/0 and other forms like 1^infinity on the wikipedia page (https://en.wikipedia.org/wiki/Indeterminate_form)
    (7 votes)
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  • old spice man green style avatar for user Taksh
    Can L'Hopital's rule be proved without assuming that f'(a) = 0 and g'(a) = 0 ?
    (3 votes)
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  • duskpin ultimate style avatar for user Lee
    Sal said it was only for the special case. So what about the proof of the general case?
    (4 votes)
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  • female robot grace style avatar for user Preslava Boycheva
    Since x is also approaching 0 (it approaches a, which in turn approaches 0), shouldn't we also substitute f(x) and g(x) with 0 in the final equation of the limit of f'(x)/g'(x) as x approaches 0? How is this a proof when it actually takes us back to our initial problem with 0/0?
    (1 vote)
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  • blobby green style avatar for user Jomana Sbehat
    how does the lim x>0 ((ln(6x))^2/(ln(2x)^2)) is =1?
    (1 vote)
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    • starky ultimate style avatar for user KLaudano
      limit of (ln(6x)^2/ln(2x)^2) as x->0
      = (limit of ln(6x)/ln(2x)) ^ 2
      = -∞/-∞

      -∞/-∞ is indeterminate so we can apply L'Hospital's Rule

      (limit of ln(6x)/ln(2x)) ^ 2
      = (limit of (1/x)/(1/x)) ^ 2
      = (limit of 1) ^ 2
      = 1 ^ 2
      = 1
      (1 vote)
  • female robot grace style avatar for user SULAGNA NANDI
    What is the proof for, if f(x) = ±∞ and g(x) = ±∞,
    lim x-->c [f'(x)/g'(x)] = lim x-->c [f(x)/g(x)]?
    (1 vote)
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  • female robot grace style avatar for user SULAGNA NANDI
    Why is this a special case of L'Hôpital's Rule? Is this just the case where f(x) = 0 and g(x) = 0, or is there something else special about this case?
    (1 vote)
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