L'hopital's Rule

L’Hopital’s Rule Adrian Down 16779577 July 20, 2005

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Motivation

Comes as a direct consequence of the Mean Value theorem. Useful for (x) computing limits of the form fg(x) where f (x), g(x) → 0. Connected to 0

(x) (x) limx→a fg0 (x) . Assume f (a) = g(a) = 0, and you want to compute limx→a fg(x) . Use the mean value theorem to try to approximate the function. As x approaches a, f (x) should be approximately f (a) + f 0 (y)(x − a). Using the mean value theorem, we know such a y exists, so f (x) = f (a) + f 0 (y)(x − a) for some a < y < x. Likewise, g(x) = g(a) + g 0 (z)(x − a) for some a < z < x. So

f 0 (y)(x − a) f 0 (y) f 0 (x) f (x) = 0 = 0 ≈ 0 g(x) g (z)(x − a) g (z) f (y) since y and z approach x as x approaches a. Keep this idea in mind for the proof.

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Proof

Theorem (generalized mean value theorem): let f, g be continuous on [a, b], and differentiable on (a, b). Then ∃x ∈ (a, b) such that f 0 (x) · (g(b) − g(a)) = g 0 (x)(f (b) − f (a)). Note: another way to think about this is that if g 0 (x) 6= 0, then f 0 (x) f (b) − f (a) = 0 g (x) g(b) − g(a) 1

This is exactly the thing that we want to take the limit of from our heuristic argument above. Will tell us that we can take y and z to be the same. Proof: same as the mean value theorem. Make a difference function, h(x) = f (x)(g(b) − g(a)) − g(x)(f (b) − f (a)). Note that it is differentiable on (a, b) and continuous on [a, b]. Do computation to get that h(a) = h(b). By Rolle’s theorem, ∃x ∈ (a, b) such that h0 (x) = 0. Note that the f (a), g(a), etc. terms in h are constants, so h0 (x) = f 0 (x)(g(b) − g(a)) − g 0 (x)(f (b) − f (a)) h0 (x) = 0 → f 0 (x)(g(b) − g(a)) = g 0 (x)(f (b) − f (a)) Note that this theorem reduces to the mean value theorem by taking g(x) = x. Theorem: let f, g be differentiable functions (on an open interval I con0 (x) taining some point a) such that g 0 is continuous on I and limx→a fg0 (x) exists is finite, and equals L ∈ R, if limx→a f (x) = limx→a g(x) = 0, then (x) limx→a fg(x) = L. Note: we need the fact that g 0 (x) 6= 0 in some area close to a. Proof: let  > 0. Can take g 0 6= 0 on I by shrinking I if necessary → g 0 is strictly increasing or strictly decreasing → g 0 is invertible on I → g 0 (x) = 0 for at most 1 x ∈ I. Shrink I so that g 6= 0 on I. The lesson is that if there could be problems on our interval, we can shrink it to a safe area since we’re taking the limit. Let K ∈ R such that L < K < L+. Want to get the quotient between L f 0 (x) f 0 (x) and K. limx→a g0 (x) = L → ∃α > 0 such that a < x < α → g0 (x) < K (taking  = K −L in the definition of the limit). If a < x < y < α → ∃z ∈ (x, y) such 0 (z) (x)−f (y) (x)−f (y) that fg(x)−g(y) = fg0 (z) by the generalized mean value theorem → fg(x)−g(y)
x→a

f (x) − f (y) ≤K g(x) − g(y)

(y) but as x approaches a, lim f (x) = lim g(x) = 0, so fg(y) ≤ K. Aside: limx→a+ f (x) = L → ∀ > 0, ∃δ > 0 such that a < x < a + δ → |f (x) − L| < . (y) ≤ K < L +  is true ∀y ∈ (a, α). Halfway to making our limit. So fg(y) (y) Similarly, ∃β such that β < y < a → fg(y) > L −  by taking L −  < K < L and using the same argument as before.

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Take δ = min{α − a, a − β}, so if 0 < |x − a| < δ → f (x) g(x)

f (x) g(x)

< L +  and

(x) | fg(x)

> L − , so − L| <  Key idea is just applying the mean value theorem to go from the ratio of quantities to a ratio of derivatives, and then using the fact that limx→a f (x) = limx→a g(x) = 0. Notes: 1) If limx→a f (x) = limx→a g(x) = ∞, then the theorem still holds 2) Can also consider limx→∞ or limx→−∞ or limx→a+ or limx→a− . . Others are 0·∞, 00 . L’Hopital’s 3) 00 is called an indeterminate form, as is ∞ ∞ rule applies to all of these. . 4)L’Hopital’s rule is NOT applicable to ∞ 0

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Examples

sin x x 1) limx→0 cosxx−1 = limx→0 − 2x = limx→0 − cos = − 12 . 2 2 2 2) limx→∞ xex = limx→∞ 2x = limx→∞ e2x = 0. ex

3) limx→0+ x · ln x = limx→0+

ln x 1 x

= limx→0+

1 x −1 x2

= limx→0+ −x = 0

4) limx→0 xx = limx→0+ ex ln x . Because ex is continuous, take the limit of the exponent separately. From the last example, it is 0. Thus the given limit is e0 = 1. 1 1 1 5) limx→∞ x x = limx→∞ e x ln x . limx→∞ x1 ln x = limx→∞ 1x = 0 1

6) limx→∞ (1+ x1 )x = limx→∞ ex ln(1+ x ) . limx→∞ x·ln(1+ x1 ) = limx→∞ limx→∞

−1 1 1 ( x2 ) 1+ x ( −1 x2 )

7) limx→0 ( ex1−1 − x1 ) = limx→0

x−ex +1 x(ex −1)

= limx→0

1−ex xex +ex +1

= limx→0

ln(1+ x1 ) 1 x

−ex xex +2ex

=

−1 2

Useful for functions that get simpler or stay the same when differentiated. Useful for polynomials, trig functions, and exponentials. Note: limx→a f (x) = +∞ → ∀M, ∃δ such that 0 < |x−a| < δ → |f (x)| > M.

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