LH opitals Rule LH opitals Rule provides a convenient way of - PDF document

L’Hˆ opital’s Rule L’Hˆ opital’s Rule provides a convenient way of finding limits of in- determinate quotients. Effectively, it states that if one wishes to find a limit of a quotient f ( x ) g ( x ) and both f ( x ) and g ( x ) either → 0 or both → ±∞ , then the limit of the quotient f ( x ) g ( x ) is equal to the limit of the quotient f ′ ( x ) g ′ ( x ) of the derivatives if the latter limit exists. Theorem 1 (L’Hˆ opital’s Rule) . Let f and g be differentiable on an open interval containing c except possibly at some fixed point c in the f ′ ( x ) interval. If lim x → c f ( x ) = lim x → c g ( x ) = 0 and lim x → c g ′ ( x ) exists, then f ( x ) lim x → c g ( x ) exists and f ( x ) f ′ ( x ) lim x → c g ( x ) = lim x → c g ′ ( x ) . This is just the most basic of literally dozens of variations. The anal- ogous result holds if the numerator and denominator both approach ±∞ , or if the limit is one-sided, or if the limit is at ∞ or at −∞ . Let’s look at some examples and then we will prove L’Hˆ opital’s Rule. The proof will depend on a generalization of the Mean Value Theorem , the General Mean Value Theorem . sin h cos h Example: lim h → 0 = lim h → 0 = 1. h 1 1 − cos h sin h Example: lim h → 0 = lim h → 0 = 0. h 1 Note both of these have been calculated before. Indeed, the reasoning here is somewhat circular, since it depends on the calculation of the derivatives of the sin and cos function, which were themselves derived using these limits. 5 x 2 +8 sin x − 1 Example: lim x →∞ x 2

2 We can also use it indirectly in cases we may think of symbolically as the following: 0 0 ∞ 0 0 · ∞ 1 ∞ We give examples of each. The general idea, in each case, is to rewrite the expression so that one winds up needing the limit of something to which one may apply L’Hˆ opital’s Rule. Example of the 0 · ∞ Case: lim x → 0 + x ln x . Here, x → 0 but ln x → −∞ . We change it into a quotient by writing x ln x = ln x 1 /x . The calculation becomes 1 /x lim x → 0 + x ln x = lim x → 0 + ln x 1 /x = lim x → 0 + − 1 /x 2 = lim x → 0 + ( − x ) = 0. Example of the 0 0 Case: lim x → 0 + x sin x . Here, we use the definition of an exponent to write x sin x as e sin x ln x . We then need only find the limit of the exponent, which brings us to the 0 · ∞ case we already dealt with. The calculations are as follows. ln x lim x → 0 + sin x ln x = lim x → 0 + 1 / sin x = sin2 x = lim x → 0 + − sin 2 x 1 /x lim x → 0 + x cos x . − cos x We can apply L’Hˆ opital’s Rule again to evaluate the latter limit! lim x → 0 + − sin 2 x 2 sin x cos x x cos x = lim x → 0 + − − x sin x +cos x = 0. So, lim x → 0 + x sin x = e 0 = 1. Example of the 1 ∞ Case: lim x →∞ (1 + 5 /x ) x . Again, we use the definition of an exponent to write (1 + 5 /x ) x = e x ln(1+5 /x ) and find the limit of the exponent. ln(1+5 /x ) lim x →∞ x ln(1 + 5 /x ) = lim x →∞ = 1 /x − 5 / [ x 2 (1+5 /x )] 5 lim x →∞ = lim x →∞ 1+5 /x = 5. − 1 /x 2 So, lim x →∞ (1 + 5 /x ) x = e 5 . The ∞ 0 Case: This may be handled like the 0 0 case. Indeed, the reciprocal of the ∞ 0 case is the 0 0 case. Summary of Special Cases • 0 · · · ∞ lim f ( x ) g ( x ), where f ( x ) → 0 and g ( x ) → ∞

3 1 /g ( x ) (giving the 0 f ( x ) g ( x ) Write f ( x ) g ( x ) as either 0 case) or 1 /f ( x ), (giving the ∞ ∞ case). • 0 0 lim f ( x ) g ( x ) , where f ( x ) → 0 + and g ( x ) → 0 Write f ( x ) g ( x ) as e g ( x ) ln f ( x ) . lim g ( x ) ln f ( x ) is then the 0 · ∞ case, since ln f ( x ) → −∞ as f ( x ) → 0 + . • 1 ∞ lim f ( x ) g ( x ) , where f ( x ) → 1 and g ( x ) → ∞ Write f ( x ) g ( x ) as e g ( x ) ln f ( x ) . lim g ( x ) ln f ( x ) is then the 0 · ∞ case, since ln f ( x ) → 0 as f ( x ) → 1. • ∞ 0 lim f ( x ) g ( x ) , where f ( x ) → ∞ and g ( x ) → 0. Write f ( x ) g ( x ) as e g ( x ) ln f ( x ) . lim g ( x ) ln f ( x ) is then the 0 · ∞ case, since ln f ( x ) → ∞ as f ( x ) → ∞ . Phony Proof of L’Hˆ opital’s Rule What’s wrong with this proof of L’Hˆ opital’s Rule? Hint: Every step, taken individually, seems correct. Assume f and g satisfy the hypotheses. Since the values of f and g at c are irrelevant, we may assume f ( c ) = g ( c ) = 0. Since we are taking a limit at c , the hypotheses imply that both f and g satisfy the hypotheses of the Mean Value Theorem as long as we are close to c . Thus: By the Mean Value Theorem, if x is relatively close to c , there is some ξ between c and x such that f ( x ) − f ( c ) = f ′ ( ξ )( x − c ). Similarly, there is some ξ between c and x such that g ( x ) − g ( c ) = g ′ ( ξ )( x − c ). It follows that f ( x ) g ( x ) = f ( x ) − f ( c ) g ( x ) − g ( c ) = f ′ ( ξ )( x − c ) g ′ ( ξ )( x − c ) = f ′ ( ξ ) g ′ ( ξ ) . Hence, f ′ ( ξ ) f ′ ( ξ ) f ( x ) lim x → c g ( x ) = lim x → c g ′ ( ξ ) = lim ξ → c g ′ ( ξ ) ,

4 since obviously ξ → c whenever x → c . The Generalized Mean Value Theorem We will prove the basic case of L’Hˆ opital’s Rule. The proof depends on a more general version of the Mean Value Theorem. We will state the Generalized Mean Value Theorem , show how it may be used to prove L’Hˆ opital’s Rule, and then prove the Generalized Mean Value Theorem. Theorem 2 (The Generalized Mean Value Theorem) . Suppose func- tions f and g are continuous on the closed interval [ a, b ] , differentiable on the open interval ( a, b ) and g ′ ( x ) is never 0 on ( a, b ) . Then there is a number c ∈ ( a, b ) such that f ( b ) − f ( a ) g ( b ) − g ( a ) = f ′ ( c ) g ′ ( c ) . Note, if g is the identity function g ( x ) = x , the Generalized Mean Value Theorem reduces to the original Mean Value Theorem. The proof will be very similar. First, we’ll use the Generalized Mean Value Theorem to prove L’Hˆ opital’s Rule. Proof. Assume f and g satisfy the hypotheses of L’Hˆ opital’s Rule. Since the values of f ( c ) and g ( c ) don’t matter, we may assume f ( c ) = g ( c ) = 0. Then, for any x near c , the hypotheses of the Generalized Mean Value Theorem will hold in the interval containing c and x and there will be some number ξ between c and x such that f ( x ) g ( x ) = f ( x ) − f ( c ) g ( x ) − g ( c ) = f ′ ( ξ ) g ′ ( ξ ) . Obviously, ξ → c as x → c , so f ( x ) f ′ ( ξ ) f ′ ( x ) lim x → c g ( x ) = lim x → c g ′ ( ξ ) = lim x → c g ′ ( x ) . � We’re now ready to prove the Generalized Mean Value Theorem. Theorem 3 (The Generalized Mean Value Theorem) . Suppose func- tions f and g are continuous on the closed interval [ a, b ] , differentiable on the open interval ( a, b ) and g ′ ( x ) is never 0 on ( a, b ) . Then there is a number c ∈ ( a, b ) such that f ( b ) − f ( a ) g ( b ) − g ( a ) = f ′ ( c ) g ′ ( c ) . Recall the proof of the Mean Value Theorem depended on a function φ ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ) which represented the distance b − a between a line and a curve. To prove the Generalized Mean Value Theorem, we use a similar function.

5 Proof. Let φ ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) g ( b ) − g ( a ) · ( g ( x ) − g ( a )). It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [ a, b ]. Thus, there must be some c ∈ ( a, b ) such that φ ′ ( c ) = 0. We first obtain φ ′ ( c ) as follows. φ ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) g ( b ) − g ( a ) · g ′ ( x ) φ ′ ( c ) = f ′ ( c ) − f ( b ) − f ( a ) g ( b ) − g ( a ) · g ′ ( c ) Since φ ′ ( c ) = 0, it follows that f ′ ( c ) − f ( b ) − f ( a ) g ( b ) − g ( a ) · g ′ ( c ) = 0 f ′ ( c ) = f ( b ) − f ( a ) g ( b ) − g ( a ) · g ′ ( c ) g ′ ( c ) = f ( b ) − f ( a ) f ′ ( c ) � g ( b ) − g ( a )