University:
California State University, Northridge
Course:
MATH 150AL | Calculus I Laboratory
Academic year:

2004
Views:

318

Pages:

14
Author:

Umnyangocj1

Indeterminacy: unsuitable breaking in limits Attention: Indeterminacy 2 3 [x + 1] ∞ [x − 1] lim 3 lim 2 x→∞ [x − 3] x→1 [x − 1] ∞ 0 0 Wrong breaking! Limit laws do not apply! lim[1 − cos x][cot x] ( 0 · ∞ ) x→0 Substitution is undefined! p p lim [ x2 + x] − [ x2 − x] ( ∞ − ∞ ) x→∞ Rewrite first! ∞ Indeterminacy : ∞ [x2 + 1] lim 3 x→∞ [x − 3] ∞ Indeterminacy : ∞ [x2 + 1] lim 3 x→∞ [x − 3] division by highest exponent of x: x2 1 + 3 0+0 3 x x lim 3 = =0 x→∞ x 1−0 3 − 3 3 x x 0 Indeterminacy : 0 3 [x − 1] lim 2 x→1 [x − 1] 0 Indeterminacy : 0 3 [x − 1] lim 2 x→1 [x − 1] factoring and re-grouping: 2 2 (x − 1)(x + x + 1) (x − 1) (x + x + 1) lim = lim · x→1 x→1 (x − 1) (x − 1)(x + 1) (x + 1) (x2 + x + 1) 3 3 (x − 1) = [lim ] · [lim ]=1· = x→1 (x − 1) x→1 (x + 1) 2 2 Indeterminacy 0 · ∞: lim[1 − cos x] · [cot x] x→0 Indeterminacy 0 · ∞: lim[1 − cos x] · [cot x] x→0 factoring, re-grouping, and special limits: cos x 1 − cos x x cos x lim(1 − cos x) = lim x→0 sin x x→0 x sin x 1 Indeterminacy 0 · ∞: lim[1 − cos x] · [cot x] x→0 factoring, re-grouping, and special limits: cos x 1 − cos x x cos x lim(1 − cos x) = lim x→0 sin x x→0 x sin x 1 1 − cos x x cos x = [lim ][lim ][lim ] = 0·1·1 = 0 x→0 x→0 sin x x→0 1 x Indeterminacy ∞ − ∞: p p lim [ x2 + x] − [ x2 − x] x→∞ Indeterminacy ∞ − ∞: p p lim [ x2 + x] − [ x2 − x] x→∞ multiplying by conjugate, re-grouping: √ √ √ √ ( x2 + x − x2 − x) ( x2 + x + x2 − x) √ √ = lim x→∞ 1 ( x2 + x + x2 − x) Indeterminacy ∞ − ∞: p p lim [ x2 + x] − [ x2 − x] x→∞ multiplying by conjugate, re-grouping: √ √ √ √ ( x2 + x − x2 − x) ( x2 + x + x2 − x) √ √ = lim x→∞ 1 ( x2 + x + x2 − x) 2 2 x +x−x +x √ = lim √ x→∞ x2 + x + x2 − x 2x √ = lim √ x→∞ x2 + x + x2 − x 2x √ = lim √ x→∞ x2 + x + x2 − x ∞/∞: division by the highest exponent = lim r x→∞ 2x xr x2 x + + x2 x2 x2 x − x2 x2 2 √ =√ =1 1+0+ 1−0

Indeterminacy: Unsuitable Breaking in Limits

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