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

2021
Views:

414

Pages:

2
Author:

Shyanne Johnson

Review: Limits at Infinity These are the techniques we had from Calculus I to find the horizontal asymptote of a function. 1. When dealing with a fraction: k (a) “ = 0” If the numerator goes to a constant, and the denominator ∞ goes to infinity, then overall the fraction goes to zero. k (b) “ = DNE” If the numerator goes to a constant, and the denom0 inator goes to 0, then overall the fraction goes to infinity (plus or minus); or you can say that it Does Not Exist (DNE). (c) Standard Algebraic Techniques: √ i. Divide by a√power of x (RECALL: If x > 0, x = x2 . If x < 0, then x = − x2 ) ii. Rationalize (usually when square roots are added/subtracted) (d) Some examples: 1 − xx2 1 x2 − x 1−0 = = = lim x→∞ 3x2 − 4 x→∞ 3 − 4 42 3 − 0 3 x √ √ √ √ √ √ x+3+ x √ lim x + 3 − x = lim x + 3 − x · √ = x→∞ x→∞ x+3+ x lim lim √ x→∞ x+3−x √ =0 x+3+ x 2. Some functions with horizontal asymptotes: (a) f (x) = tan−1 (x) has horizontal asymptotes at y = π2 (as x → ∞), and −π 2 (as x → −∞) 1 (b) f (x) = p has a horizontal asymptote at y = 0 if p ≥ 1 and x → ∞. x 3. Using l’Hospital’s Rule: The Rule: If you have “ 00 ” or “ ±∞ ±∞ ”, then f (x) f 0 (x) = lim 0 x→∞ g(x) x→∞ g (x) lim You may not be given a fraction- we may have to do some algebra first! Sometimes you can use l’Hospital’s rule for a product, f (x) · g(x), and sometimes for exponentiation, (f (x))g(x) . See the examples below. NOTE: f (x)g(x) = eg(x)·ln(f (x)) 1 4. Examples 1 (a) lim x x x→∞ 1 1 First, some algebra: x x = e x ln(x) Now, we need the limit of rule: ln(x) x , which we can get using l’Hospital’s 1 ln(x) = lim x = 0 x→∞ 1 x→∞ x lim so overall: 1 1 lim x x = lim e x ln(x) = elimx→∞ x→∞ ln(x) x x→∞ = e0 = 1 (b) lim x2 e−x x→∞ Rewrite the product as a fraction so we can apply l’Hospital’s rule: x2 2x 2 = lim x = lim x = 0 x→∞ ex x→∞ e x→∞ e lim x2 e−x = lim x→∞ (c) lim ln(x + 1) − ln(x) x→∞ Rewrite using the rules of logarithms: x+1 x+1 = ln lim = ln(1) = 0 lim ln(x+1)−ln(x) = lim ln x→∞ x→∞ x→∞ x x (d) lim (1 + 1/x)x x→∞ Rewrite using the exponential function: (1 + 1/x)x = ex ln(1+1/x) Then take the limit of this exponent: lim x ln(1 + 1/x) = lim x→∞ x→∞ ln(1 + 1/x) 1 x 1 −1 1+1/x · x2 −1 x→∞ x2 = lim so overall, lim (1 + 1/x)x = lim ex ln(1+1/x) = e1 = e x→∞ x→∞ √ (e) lim x→−∞ x2 + 3 3x − 2 √ lim x→−∞ q 1 − 1 + x32 1 x2 + 3 − √x2 =− · 1 = lim 2 x→−∞ 3 3x − 2 3− x x 2 =1

Review Limits at Infinity - Study Guide

Related Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock