MATH241 Lecture Notes - Lecture 9: Coefficient, Tangent
9 May 2018
School
Department
Course
Professor
MATH241 - Lecture 9 - Limits and Derivatives
2.6: Limits at Infinity (continued)
Example
:
Find the vertical and horizontal asymptotes of g( ) = xx + 1
√4x + 2
2
Solution:
Vertical Asymptote (VA):
= -1 is a candidate for a vertical asymptote because g( ) is undefined due to division byx x
ml = =
x + 1
√4x + 2
2√6
0 −− ∞
x→ − 1 −
ml = =
x + 1
√4x + 2
2√6
0 +− ∞
x→ − 1−
Thus g( ) has a vertical asymptote at = -1x x
Horizontal Asymptote (HA):
ml =
x + 1
√4x + 2
2
ml (factor out largest power of x in denominator)
x
√x 2
(1 + x
1
√4 + 2
x2)
x→ ∞ x→ ∞
= ml ( cancels out)
x
x
(1 + x
1
√4 + 2
x2)x
x
x→ ∞
= ml 1 + x
1
√4 + 2
x2
x→ ∞
= 1 + 0
√4 + 0
= = 2
1
2
ml =
x + 1
√4x + 2
2
ml (bring out largest power of x in denominator)
x
√x 2
(1 + x
1
√4 + 2
x2)
x→ − ∞ x→ − ∞
= ml x
x
| |
(1 + x
1
√4 + 2
x2)
(replace with opposite of x since it’s going to )x→ − ∞ − ∞
x
| | =x x ≥ 0
x x < 0
= ml x
(−x)
(1 + x
1
√4 + 2
x2)
x→ − ∞
= ml (-1) (cancel but leave -1 behind)
(1 + x
1
√4 + 2
x2)x
(−x)
x→ − ∞
= (-1) 1 + 0
√4 + 10
= -2
Thus g( ) has a horizontal asymptote at and another at x y = 2 y= − 2
Infinite Limits at Infinity
Definition: A limit f( ) = means that f( ) gets larger in magnitude as gets larger inml x ± ∞ x x
±x→ ∞
magnitude.
Examples
:
Example
:
Determine the limits:
A) ml x + 4x
2
x+ 4x + 2
3 3
x→ ∞
B) ml x4−x3
x→ ∞
Document Summary
Math241 - lecture 9 - limits and derivatives. Find the vertical and horizontal asymptotes of g( = -1 is a candidate for a vertical asymptote because g( x. ) is undefined due to division by ml x 1 ml x 1 . 1 (factor out largest power of x in denominator) ml x ml x . 4 + 2 ( 1 + x x2) x. X 2 x ml x . 1 (bring out largest power of x in denominator) (replace with opposite of x since it"s going to x| | = x x (cancel ( x) x but leave -1 behind) x2) ( 1 + x. 1 x 0 x < 0 ( 1 + x x2) 4 + 2 ( 1 + x x2) = ml x| x x . = ml x ( x) x. ) has a horizontal asymptote at y = 2 and another at y = 2.
