Limits Involving Infinity

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1f x

x

1lim 0x x

As the denominator gets larger, the value of the fraction gets smaller.

There is a horizontal asymptote if:

limx

f x b

or limx

f x b

2lim

1x

x

x

Example 1:

2limx

x

x

This number becomes insignificant as .x

limx

x

x 1

There is a horizontal asymptote at 1.

sin xf x

x

Example 2:

sinlimx

x

x Find:

When we graph this function, the limit appears to be zero.1 sin 1x

so for :0x 1 sin 1x

x x x

1 sin 1lim lim limx x x

x

x x x

sin0 lim 0

x

x

x

by the sandwich theorem:

sinlim 0x

x

x

Example 3: 5 sinlimx

x x

x

Find:

5 sinlimx

x x

x x

sinlim 5 limx x

x

x

5 0

5

Infinite Limits:

1f x

x

0

1limx x

As the denominator approaches zero, the value of the fraction gets very large.

If the denominator is positive then the fraction is positive.

0

1limx x

If the denominator is negative then the fraction is negative.

vertical asymptote at x=0.

Example 4:

20

1limx x

20

1limx x

The denominator is positive in both cases, so the limit is the same.

20

1 limx x

End Behavior Models:

End behavior models model the behavior of a function as x approaches infinity or negative infinity.

A function g is:

a right end behavior model for f if and only if

lim 1x

f x

g x

a left end behavior model for f if and only if

lim 1x

f x

g x

Test ofmodel

Our modelis correct.

xf x x e Example 7:

limx

x

x e

x

lim1

x

x

e

x

1 0 1

limx

xx

x e

e

lim 1xx

x

e 0 1 1

Show that g(x) = x is a rightend behavior model for f(x).

Show that h(x) = e-x is a leftend behavior model for f(x).

1

2

3

x

x

f x x

f x e

f x x e

xf x x e Example 7:

g x x becomes a right-end behavior model.

xh x e becomes a left-end behavior model.

10 10x

1.43 11.43y

Use:

On your calculator, graph:

5 4 2

2

2 1

3 5 7

x x xf x

x x

Example 7:

End behavior models give us:

5

2

2

3

x

x

32

3

x

dominant terms in numerator and denominator

p

Power function forthe end behavior model!

Often you can just “think through” limits using the following.

1lim sin

x x 0 0

1lim sin lim sin( ) 0

1

x xx

xp

𝐥𝐢𝐦𝒙→∞

𝒇 (𝒙 )= 𝒍𝒊𝒎𝒙→𝟎+¿ 𝒇 (𝟏𝒙 )

¿

𝐥𝐢𝐦𝒙→−∞

𝒇 (𝒙 )=𝒍𝒊𝒎𝒙→𝟎−

𝒇 (𝟏𝒙 )