University:
Whitman College
Course:
MATH-126 | Calculus II
Academic year:

2024
Views:

73

Pages:

3
Author:

Jamal Simon

Extra Practice with Limits 1. Compute each limit (algebraically) if it exists: (a) lim √1 x+h − √1 x h x −4 (b) lim 1 1 x→2 −2 x h→0 2 x2 − 2x − 3 (c) lim x→2 x−3 √ (d) lim ( x2 + 2x − x) x→∞ 3x + 2 (e) lim √ 2 x→∞ x −1 2. Determine if the sequence converges, and if it does, find the limit: (a) an = e1/n 2n+3 (b) an = n 3 2n (c) an = n! (d) {1, 0, −1, 0, 1, 0, −1, · · ·} (e) an = (n + 1)(2n + 1) n2 + 2n + 1 3. Find the limit: n X 2 2i lim 3 +1 n→∞ n i=1 n Hint: Expand the sum, then use the sum formulas to get an expression involving n only (with no sums). 1 SOLUTIONS: 1. Compute each limit (algebraically) if it exists: √1 x+h (a) lim − √1 x h h→0 lim Simplify the expression first (common denominator): √1 x+h h→0 − h √1 x √ √ √ √ 1 x− x+h x+ x+h √ = lim · √ √ ·√ = h→0 h x x+h x+ x+h x − (x + h) −1 −1 1 √ √ = 3/2 ·√ √ = √ h→0 h x·2 x 2x x x + h( x + x + h) lim x2 − 4 Simplify the denominator then simplify before taking the limit: x→2 1 − 1 x 2 (b) lim x2 − 4 (x + 2)(x − 2)2x x2 − 4 = lim = lim = lim −(x + 2)2x = −16 x→2 2−x x→2 x→2 x→2 1 − 1 2−x x 2 2x lim x2 − 2x − 3 x→2 x−3 (c) lim x2 − 2x − 3 4−4−3 = =3 x→2 x−3 2−3 (Be sure to check the numbers before doing any algebra!) √ (d) x→∞ lim ( x2 + 2x − x) Multiply by the conjugate: lim lim x→∞ √ √ (x2 + 2x) − x2 2x x2 + 2x + x x2 + 2x − x · √ 2 = x→∞ lim √ 2 = x→∞ lim √ 2 = x + 2x + x x + 2x + x x + 2x + x 2x/x 2 2 2 lim √ 2 = lim q 2 = lim q = =1 x→∞ ( x + 2x + x)/x x→∞ x +2x 2 + 1 x→∞ 1 + x2 + 1 x2 3x + 2 Divide numerator and denominator by x: (e) x→∞ lim √ 2 x −1 3 + x2 3 + x2 3 + x2 3x + 2 q q √ √ lim = lim = lim = x→∞ lim =3 x→∞ x2 − 1 x→∞ ( x2 − 1)/x x→∞ x2 −1 1 − 12 2 x 2. Determine if the sequence converges, and if it does, find the limit: (a) an = e1/n SOLUTION: The exponential function is continuous at zero, so: lim e1/n = elimn→∞ (1/n) = e0 = 1 n→∞ 2 x 2n+3 3n SOLUTION: 0 < 2/3 < 1, so (2/3)n → 0 as n → ∞. Therefore: (b) an = n 2n 23 2 2n+3 = lim = 8 lim n→∞ 3n n→∞ 3 n→∞ 3n lim =0 2n n! SOLUTION: If we examine the sequence term by term: (c) an = a1 = a2 = a3 = a4 = 2 1 2·2 1·2 ⇒ 2·2·2 1·2·3 an+1 = 2 an n+1 2·2·2·2 1·2·3·4 And we see that an ≤ 2 for all n. We could use a kind of squeeze theorem: 0 ≤ an+1 = 2 2 an ≤ ·2→0 n+1 n+1 (d) {1, 0, −1, 0, 1, 0, −1, · · ·} SOLUTION: The limit does not exist. (n + 1)(2n + 1) (e) an = n2 + 2n + 1 SOLUTION: Use either l’Hospital’s rule or divide numerator and denominator by n2 to get the limit is 2. 3. Find the limit: n X 2 2i lim 3 +1 n→∞ n i=1 n Hint: Expand the sum, then use the sum formulas to get an expression involving n only (with no sums). SOLUTION: Using the hint, we’ll do the algebra first: n n n X 2X 2 2i 12i 2 12 X 12 n(n + 1) 2 6(n + 1) 3 +1 = + = i+ 1= 2 + ·n = +2 2 2 n n n i=1 n i=1 n 2 n n i=1 n i=1 n n X Finally, taking the limit as n → ∞, we get 6 + 2 = 8. 3

Math 126 Extra Practice with Limits

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