MAT137 Calculus! Lecture 19 Today: L Hôpital s Rule 11.5 The Indeterminate Form (0/0) 11.6 The Indeterminate Form ( / ) + other Indeterminate Forms Test 2: Friday, Nov. 25. If you have a conflict, let us know by this Friday, Nov. 18. Next: 4.1 Mean Value Thm. official website http://uoft.me/mat137
Compute x 0 x 2 2x e x 1. y y y = e x 1 y = x y = x y = e x 1 x y = x 2 2x x y = 2x Near x = 0, we have x 2 2x e x 1 2x x. y = 2x y = x 2 2x
Theorem (L Hôpital s Rule) IF THEN f (x) x a g(x) is of the indeterminate form 0 0 or f and g are differentiable near a (except possibly at a), g and g are not 0 near a (except possibly at a), f (x) x a g (x) exists or is ±, f (x) x a g(x) = f (x) x a g (x) The conclusion also holds if x as long as f and g are differentiable on some interval (N, ) on which g(x) and g (x) are not 0. It also holds for x a +, x a, x.
L Hôpital s Rule (Baby Version) IF THEN f (x) x a g(x) is of the indeterminate form 0 0 f and g are differentiable on an interval containing a, f (a) = g(a) = 0 f and g are continuous g (a) 0 f (x) x a g(x) = f (a) g (a) = f (x) x a g (x)
Example 1 Compute x 0 x 2 2x e x 1.
Repeated Example 2 x 2 Compute x e x. Homework For that for any k N, x k x e x = 0 Note: This means that e x approaches infinity faster than x k (or any polynomial) as x.
Careless Bob, again! Example 3 sin x Compute x π 1 cos x. Solution: x π sin x 1 cos x L H = x π cos x sin x =
Indeterminate Products 0 IF x a f (x) = 0 and x a g(x) = ±, THEN f (x)g(x) =??? x a 1 If f (x) = x and g(x) = 1/x 2, then f (x) = 0, g(x) =, and f (x)g(x) = 1/x DNE. x 0 x 0 x 0 x 0 2 If f (x) = x 2 and g(x) = 1/x 2, then f (x) = 0, g(x) =, and f (x)g(x) = 1 = 1. x 0 x 0 x 0 x 0 3 If f (x) = x 3 and g(x) = 1/x 2, then f (x) = 0, g(x) =, and f (x)g(x) = x = 0. x 0 x 0 x 0 x 0 But we can t use L Hôpital s Rule... at least not yet!
Indeterminate Products 0 If f (x) = 0 and g(x) = ±, we write the product f (x)g(x) as a x a x a quotient So we have f (x) x a 1 g(x) g(x) x a 1 f (x) is of type 0/0 fg = f 1 g is of type / and L Hôpital s Rule applies now or fg = g 1 f
Example 4 Compute x ln x. x 0 +
Indeterminate Differences If x a f (x) = and x a g(x) =, then [f (x) g(x)] x a is an indeterminate form of type.
Example 5 Compute x 0 (csc x cot x).
Example 6 Compute [x x 2 + x]. x Which solution is right? Solution 1 [x x 2 + x] = x Solution 2 [x x 2 + x] [x + x 2 + x] x [x + x 2 + x] x 2 (x 2 + x) = x [x + x 2 + x] = x x x [1 + 1 + 1 x ] = x 1 = 1 [1 + 1 + 1 x ] 2 [x x 2 + x] = ( ) = x
Indeterminate Forms for Limits Which of the following are indeterminate forms for its? 1 1 2 0 3 0 4 0
Indeterminate and Non-Indeterminate Forms Indeterminate Forms for Limits 0 0 0 0 0 1 0 Non-Indeterminate Forms for Limits 1 A it in any of these forms is equal to 0: 1 0 0 2 A it in any of these forms is : 0 + + 1
Example 7 Compute (1 + 1 x x x ).
Indeterminate Powers: 0 0, 1, 0 When x a [f (x)] g(x) has the indeterminate form 1, 0 0 or 0, we can logarithms to transform exponentiation into a product. Let y = [f (x)] g(x), then ln y = g(x) ln f (x). Since e x is continuous y = x a x a eln y = e ln y x a. In all the three cases above, this method leads to computing g(x) ln f (x) which is always indeterminate of type 0. x a
Example 8 Compute x 0 + x x. Answer: 1 1 2 0 3 e
Example 9 3x + cos x Compute. x x Answer: 1 DNE 2 3 3 0
Example 9 3x + cos x Compute. x x Answer: 1 DNE 2 3 3 0
Example 9 3x + cos x Compute. x x Note that this it has the form /. Wrong Solution Applying L Hôpital s Rule we have 3x + cos x x x L H? = x 3 sin x 1 DNE L Hôpital s Rule doesn t apply!!! f (x) To reach any conclusion from L Hôpital s Rule x a g (x) ±. must exist or be
Example 9 3x + cos x Compute. x x Right Solution Note that 3x + cos x x x So by the Squeeze Theorem Hence, x 3x + cos x x = (3 + cos x ) x x 0 cos x 1 0 cos x 1 x x = 3. cos x x x = 0