Lecture 7 : Indeterminate Forms Recall that we calculated the following limit using geometry in Calculus 1: sin x lim = 1. x→0 x

0 Definition An indeterminate form of the type 0 is a limit of a quotient where both numerator and denominator approach 0. Example ex − 1 x−2 cos x lim lim lim π x→0 x→∞ −x x→ π sin x e 2 x − 2

∞ f(x) Definition An indeterminate form of the type ∞ is a limit of a quotient g(x) where f(x) → ∞ or −∞ and g(x) → ∞ or −∞ . Example x2 + 2x + 1 x−1 lim lim . x→∞ ex x→0+ ln x

1 L’Hospital’s Rule Suppose lim stands for any one of

lim lim lim lim lim x→a x→a+ x→a− x→∞ x→−∞

f(x) 0 ∞ and g(x) is an indeterminate form of type 0 or ∞ .

f 0(x) If lim g0(x) is a finite number L or is ±∞, then

f(x) f 0(x) lim = lim . g(x) g0(x)

(Assuming that f(x) and g(x) are both differentiable in some open interval around a or ∞ (as appro- priate) except possible at a, and that g0(x) 6= 0 in that interval).

Definition lim f(x)g(x) is an indeterminate form of the type 0 · ∞ if

lim f(x) = 0 and lim g(x) = ±∞.

Example limx→∞ x tan(1/x)

0 We can convert the above indeterminate form to an indeterminate form of type 0 by writing f(x) f(x)g(x) = 1/g(x)

∞ or to an indeterminate form of the type ∞ by writing g(x) f(x)g(x) = . 1/f(x)

We them apply L’Hospital’s rule to the limit.

2 Indeterminate Forms of the type 00, ∞0, 1∞. Type Limit

00 lim [f(x)]g(x) lim f(x) = 0 lim g(x) = 0

∞0 lim [f(x)]g(x) lim f(x) = ∞ lim g(x) = 0

1∞ lim [f(x)]g(x) lim f(x) = 1 lim g(x) = ∞

2 1 Example limx→0(1 + x ) x .

Method 1. Look at lim ln[f(x)]g(x) = lim g(x) ln[f(x)].

2. Use L’Hospital to find lim g(x) ln[f(x)] = α.(α might be finite or ±∞ here. )

3. Then limf(x)g(x) = lim eln[f(x)]g(x) = eα since ex is a continuous function. (where e∞ should be interpreted as ∞ and e−∞ should be interpreted as 0. )

Indeterminate Forms of the type ∞ − ∞ occur when we encounter a limit of the form lim(f(x) − g(x)) where lim f(x) = lim g(x) = ∞ or lim f(x) = lim g(x) = −∞

1 1 Example limx→0+ x − sin x

To deal with these limits, we try to convert to the previous indeterminate forms by adding fractions etc...

3 Try these Extra Fun Examples over Lunch 2x lim x→−∞ 1 sin( x ) ln x lim x→0+ csc x

lim x tan(1/x) x→∞

2x 1 lim (e − 1) ln x x→0+

1 lim(x) x−1 x→1

4 Lecture 7 : Indeterminate Forms 2x lim x→−∞ 1 sin( x )

(Note: You could use the sandwich theorem from Calc 1 for this if you prefer.) 0 This is an indeterminate form of type 0 . By L’Hospitals rule it equals: (ln 2)2x ln 2 2x lim = lim lim x→−∞ 1 1 x→−∞ 1 x→−∞ 1 − x2 cos( x ) cos( x ) − x2 −x2 = (ln 2) lim x→−∞ 2−x Applying L’Hospital again, we get that this equals −2x (ln 2) lim x→−∞ −(ln 2)2−x

Applying L’Hospital a third time, we get that this equals

2(ln 2) 1 lim = 0 ln 2 x→−∞ −(ln 2)2−x

ln x lim x→0+ csc x

∞ This is an indeterminate form of type ∞ Applying L’Hospital’s rule we get that it equals

1/x − sin2 x lim − cos x = lim x→0+ x→0+ sin2 x x cos x − sin2 x 1 − sin2 x = lim lim = lim x→0+ x x→0+ cos x x→0+ x We can apply L’Hospital’s rule again to get that the above limit equals −2 sin x cos x lim = 0 x→0+ 1

lim x tan(1/x) x→∞

0 Rearranging this, we get an indeterminate form of type 0

−1 2 tan(1/x) 2 sec (1/x) lim = lim x x→∞ x→∞ −1 1/x x2 1 = lim = 1 x→∞ cos2(1/x)

5 2x 1 lim (e − 1) ln x x→0+ is an indeterminate form of type 00. Using continuity of the exponential function, we get 1 1 2x ln x ) 1 2x 2x lim + ln((e −1) lim + ln(e −1) limx→0+ (e − 1) ln x = e x→0 = e x→0 ln x For ln(e2x − 1) lim x→0+ ln x we apply L’Hospital to get:

2x 2x (2e ) 2x ln(e − 1) e2x−1 (2xe ) lim = lim 1 = lim x→0+ x→0+ x→0+ 2x ln x x e − 1 We apply L’Hospital again to get :

(2(e2x + 2xe2x)) = lim = 1 x→0+ 2e2x Substiuting this into the original limit, we get

2x 1 1 lim (e − 1) ln x = e = e x→0+

1 lim(x) x−1 x→1

1 1 1 lim ln(x x−1 ) lim ln(x x−1 ) lim(x) x−1 = e x→1 = e x→1 x→1 Focusing on the power we get 1 ln x lim ln(x x−1 ) = lim x→1 x→1 x − 1 0 This is an indeterminate form of type 0 so we can apply L’Hospital’s rule to get ln x 1/x lim = lim = 1 x→1 x − 1 x→1 1 Substitution this for the power of e above we get

1 1 lim(x) x−1 = e = e x→1 √ lim x2 + ln x − x x→∞ This is an indeterminate form of type ∞ − ∞ √ √ x2 + ln x + x x2 + ln x − x2 = lim ( x2 + ln x − x)√ = lim √ x→∞ x2 + ln x + x x→∞ x2 + ln x + x ln x = lim √ x→∞ x2 + ln x + x

6 ∞ This in an indeterminate form of type ∞ . We can apply L’Hospital to get ln x 1/x √ lim = lim 2x+1/x x→∞ x2 + ln x + x x→∞ √ + 1 2 x2+ln x We calculate 2x + 1/x lim √ x→∞ 2 x2 + ln x by dividing the numerator and denominator by x to get

2x + 1/x 2 + 1/x2 lim √ = lim q x→∞ 2 x2 + ln x x→∞ ln x 2 1 + x2

Applying L’Hospital to get lim ln x/x2 = lim 1/2x2 = 0, we get

2x + 1/x lim √ = 1 x→∞ 2 x2 + ln x and using this, we get ln x 1/x 0 √ lim = lim 2x+1/x = = 0 x→∞ x2 + ln x + x x→∞ √ + 1 2 2 x2+ln x Now this gives √ ln x lim x2 + ln x − x = lim √ = 0 x→∞ x→∞ x2 + ln x + x

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