Calculus (Math 1A) Lecture 10 Vivek Shende September 15, 2017 - - PowerPoint PPT Presentation

Calculus (Math 1A) Lecture 10

Vivek Shende September 15, 2017

Hello and welcome to class!

Hello and welcome to class!

Last time

Hello and welcome to class!

Last time

We discussed continuity.

Hello and welcome to class!

Last time

We discussed continuity.

Today

Hello and welcome to class!

Last time

We discussed continuity.

Today

We will define limits at infinity, and give examples of their computation.

Asymptotes of a graph

Asymptotes of a graph

Asymptotes of a graph

Asymptotes are lines which the graph approaches.

Vertical asymptotes

Vertical asymptotes

A vertical asymptote is a line x = a where lim

x→a+ f (x) = ±∞

• r

lim

x→a− f (x) = ±∞

Horizontal asymptotes

Horizontal asymptotes

A horizontal asymptote is a line y = b where lim

x→∞ f (x) = b

• r

lim

x→−∞ f (x) = b

Horizontal asymptotes

A horizontal asymptote is a line y = b where lim

x→∞ f (x) = b

• r

lim

x→−∞ f (x) = b

We’ll say what these limits mean shortly.

More asymptotes

More asymptotes

More asymptotes

More asymptotes

We won’t discuss these in this class.

Why asymptotes?

Why asymptotes?

Asymptotes know qualitative and large-scale behavior.

Why asymptotes?

Asymptotes know qualitative and large-scale behavior. Vertical asymptotes tell you where the function gets large.

Why asymptotes?

Asymptotes know qualitative and large-scale behavior. Vertical asymptotes tell you where the function gets large. Horizontal asymptotes tell you what happens when x gets large.

Why asymptotes?

Asymptotes know qualitative and large-scale behavior. Vertical asymptotes tell you where the function gets large. Horizontal asymptotes tell you what happens when x gets large. You can also learn about roots, using intermediate value theorem:

Limits at infinity

Limits at infinity

lim

x→∞ f (x) = y

Limits at infinity

lim

x→∞ f (x) = y

Informally: f (x) approaches y as x approaches infinity.

Limits at infinity

lim

x→∞ f (x) = y

Informally: f (x) approaches y as x approaches infinity. More precisely: f (x) can be brought within any fixed distance of y by ensuring that x is sufficiently large.

Limits at infinity

lim

x→∞ f (x) = y

Informally: f (x) approaches y as x approaches infinity. More precisely: f (x) can be brought within any fixed distance of y by ensuring that x is sufficiently large. In symbols:

Limits at infinity

lim

x→∞ f (x) = y

Informally: f (x) approaches y as x approaches infinity. More precisely: f (x) can be brought within any fixed distance of y by ensuring that x is sufficiently large. In symbols: for every ǫ > 0,

Limits at infinity

lim

x→∞ f (x) = y

Informally: f (x) approaches y as x approaches infinity. More precisely: f (x) can be brought within any fixed distance of y by ensuring that x is sufficiently large. In symbols: for every ǫ > 0, there exists some N > 0 such that if x > N then |f (x) − y| < ǫ.

Infinite limits at infinity

Infinite limits at infinity

lim

x→∞ f (x) = ∞

Infinite limits at infinity

lim

x→∞ f (x) = ∞

Informally: f (x) approaches infinity as x approaches infinity.

Infinite limits at infinity

lim

x→∞ f (x) = ∞

Informally: f (x) approaches infinity as x approaches infinity. More precisely: f (x) can be made arbitrarily large by ensuring that x is sufficiently large.

Infinite limits at infinity

lim

x→∞ f (x) = ∞

Informally: f (x) approaches infinity as x approaches infinity. More precisely: f (x) can be made arbitrarily large by ensuring that x is sufficiently large. In symbols:

Infinite limits at infinity

lim

x→∞ f (x) = ∞

Informally: f (x) approaches infinity as x approaches infinity. More precisely: f (x) can be made arbitrarily large by ensuring that x is sufficiently large. In symbols: for every M > 0,

Infinite limits at infinity

lim

x→∞ f (x) = ∞

Informally: f (x) approaches infinity as x approaches infinity. More precisely: f (x) can be made arbitrarily large by ensuring that x is sufficiently large. In symbols: for every M > 0, there exists some N > 0 such that if x > N then f (x) > M.

Limits at infinity from one-sided limits

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim y→0+ f (1/y)

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim y→0+ f (1/y)

Asserting the left hand side is equal to some value c is saying:

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim y→0+ f (1/y)

Asserting the left hand side is equal to some value c is saying:

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim y→0+ f (1/y)

Asserting the left hand side is equal to some value c is saying: for every ǫ, there’s a N such that x > N implies |f (x) − c| < ǫ.

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim y→0+ f (1/y)

Asserting the left hand side is equal to some value c is saying: for every ǫ, there’s a N such that x > N implies |f (x) − c| < ǫ. Asserting the left hand side is equal to some value c is saying:

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim y→0+ f (1/y)

Asserting the left hand side is equal to some value c is saying: for every ǫ, there’s a N such that x > N implies |f (x) − c| < ǫ. Asserting the left hand side is equal to some value c is saying:

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim y→0+ f (1/y)

Asserting the left hand side is equal to some value c is saying: for every ǫ, there’s a N such that x > N implies |f (x) − c| < ǫ. Asserting the left hand side is equal to some value c is saying: for every ǫ, there’s a δ such that 0 < y < δ implies |f (1/y) − c| < ǫ.

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim y→0+ f (1/y)

Asserting the left hand side is equal to some value c is saying: for every ǫ, there’s a N such that x > N implies |f (x) − c| < ǫ. Asserting the left hand side is equal to some value c is saying: for every ǫ, there’s a δ such that 0 < y < δ implies |f (1/y) − c| < ǫ. To go between one and the other, take y = 1/x and δ = 1/N.

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

It follows that the limit laws we learned before hold for limits at infinity, except the ones involving plugging in the value.

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

It follows that the limit laws we learned before hold for limits at infinity, except the ones involving plugging in the value. Also, for n > 0:

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

It follows that the limit laws we learned before hold for limits at infinity, except the ones involving plugging in the value. Also, for n > 0: we have lim

x→∞ x−n

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

It follows that the limit laws we learned before hold for limits at infinity, except the ones involving plugging in the value. Also, for n > 0: we have lim

x→∞ x−n = lim x→0+ xn

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

It follows that the limit laws we learned before hold for limits at infinity, except the ones involving plugging in the value. Also, for n > 0: we have lim

x→∞ x−n = lim x→0+ xn = 0

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

It follows that the limit laws we learned before hold for limits at infinity, except the ones involving plugging in the value. Also, for n > 0: we have lim

x→∞ x−n = lim x→0+ xn = 0

and lim

x→∞ xn

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

It follows that the limit laws we learned before hold for limits at infinity, except the ones involving plugging in the value. Also, for n > 0: we have lim

x→∞ x−n = lim x→0+ xn = 0

and lim

x→∞ xn = lim x→0+ x−n

Limits at infinity from one-sided limits

By comparing definitions, one has: lim

x→∞ f (x) = lim x→0+ f (1/x)

It follows that the limit laws we learned before hold for limits at infinity, except the ones involving plugging in the value. Also, for n > 0: we have lim

x→∞ x−n = lim x→0+ xn = 0

and lim

x→∞ xn = lim x→0+ x−n = ∞

Limits at infinity of rational functions

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = ?

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = ? We can’t directly apply the quotient law, since the top and bottom both are going to infinity as x → ∞.

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = ? We can’t directly apply the quotient law, since the top and bottom both are going to infinity as x → ∞. Instead: lim

x→∞

x2 + 2x + 3 x2 − 1 = lim

x→∞

x2 + 2x + 3 x2 − 1 · x−2 x−2

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = ? We can’t directly apply the quotient law, since the top and bottom both are going to infinity as x → ∞. Instead: lim

x→∞

x2 + 2x + 3 x2 − 1 = lim

x→∞

x2 + 2x + 3 x2 − 1 · x−2 x−2 = lim

x→∞

1 + 2x−1 + 3x−2 1 − x−2

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = lim

x→∞

1 + 2x−1 + 3x−2 1 − x−2

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = lim

x→∞

1 + 2x−1 + 3x−2 1 − x−2 Now we can apply the quotient law, because: lim

x→∞(1+2x−1+3x−2) = lim x→∞ 1+ lim x→∞ 2x−1+ lim x→∞ 3x−2 = 1+0+0 = 1

and similarly lim 1 − x−2 = 1.

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = lim

x→∞

1 + 2x−1 + 3x−2 1 − x−2 Now we can apply the quotient law, because: lim

x→∞(1+2x−1+3x−2) = lim x→∞ 1+ lim x→∞ 2x−1+ lim x→∞ 3x−2 = 1+0+0 = 1

and similarly lim 1 − x−2 = 1. Collecting these together:

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = lim

x→∞

1 + 2x−1 + 3x−2 1 − x−2 Now we can apply the quotient law, because: lim

x→∞(1+2x−1+3x−2) = lim x→∞ 1+ lim x→∞ 2x−1+ lim x→∞ 3x−2 = 1+0+0 = 1

and similarly lim 1 − x−2 = 1. Collecting these together: lim

x→∞

1 + 2x−1 + 3x−2 1 − x−2 = lim

x→∞ 1 + 2x−1 + 3x−2

lim 1 − x−2 = 1 1 = 1

Limits at infinity of rational functions

lim

x→∞

x2 + 2x + 3 x2 − 1 = 1

Try it yourself:

lim

x→∞

3x4 + x2 + 2 5x4 + 3x3 + 2x + 1 = ?

Limits at infinity of rational functions

Limits at infinity of rational functions

lim

x→∞

x + 2 x2 − 1 = ?

Limits at infinity of rational functions

lim

x→∞

x + 2 x2 − 1 = ? This time: lim

x→∞

x + 2 x2 − 1 = lim

x→∞

x + 2 x2 − 1 · x−2 x−2 = lim

x→∞

x−1 + 2x−2 1 − x−2

Limits at infinity of rational functions

lim

x→∞

x + 2 x2 − 1 = ? This time: lim

x→∞

x + 2 x2 − 1 = lim

x→∞

x + 2 x2 − 1 · x−2 x−2 = lim

x→∞

x−1 + 2x−2 1 − x−2 = lim

x→∞ x−1 + 2x−2

lim

x→∞ 1 − x−2

= 0 1 = 0

Limits at infinity of rational functions

Limits at infinity of rational functions

lim

x→∞

x2 − 1 x + 2 = ?

Limits at infinity of rational functions

lim

x→∞

x2 − 1 x + 2 = ? This time: lim

x→∞

x2 − 1 x + 2 = lim

x→∞

x2 − 1 x + 2 · x−1 x−1 = lim

x→∞

x(1 − x−1) 1 + 2x−1

Limits at infinity of rational functions

lim

x→∞

x2 − 1 x + 2 = ? This time: lim

x→∞

x2 − 1 x + 2 = lim

x→∞

x2 − 1 x + 2 · x−1 x−1 = lim

x→∞

x(1 − x−1) 1 + 2x−1

?

= ( lim

x→∞ x)

• lim

x→∞

1 − x−1 1 + 2x−1

• ?

= ∞ · 1 ? = ∞

Limit laws for possibly infinite limits

If lim

x→a g(x) exists and is finite, then:

lim

x→a[f (x) + g(x)]

= [ lim

x→a f (x)] + [ lim x→a g(x)]

lim

x→a[f (x) − g(x)]

= [ lim

x→a f (x)] − [ lim x→a g(x)]

Here, if k is a nonzero constant, we understand ∞ ± k = ∞ and −∞ ± k = −∞.

Limit laws for possibly infinite limits

If lim

x→a g(x) exists and is finite, then:

lim

x→a[f (x) + g(x)]

= [ lim

x→a f (x)] + [ lim x→a g(x)]

lim

x→a[f (x) − g(x)]

= [ lim

x→a f (x)] − [ lim x→a g(x)]

Here, if k is a nonzero constant, we understand ∞ ± k = ∞ and −∞ ± k = −∞. What is being asserted is that the limit on the left exists if and

• nly if the limit on the right exists, and in this case they are equal.

Limit laws for possibly infinite limits

If lim

x→a g(x) exists and is finite and nonzero,

lim

x→a[f (x)g(x)]

= [ lim

x→a f (x)][ lim x→a g(x)]

lim

x→a[f (x)/g(x)]

= [ lim

x→a f (x)]/[ lim x→a g(x)]

Here, if k is a nonzero constant, we understand k · ∞ = ∞ for k > 0 and k · ∞ = −∞ for k < 0.

Limit laws for infinite limits

Finally, if lim

x→a f (x) = ∞ = lim x→a g(x), then

lim

x→a f (x) + g(x) = ∞

lim

x→a f (x)g(x) = ∞

Limits at infinity of rational functions

lim

x→∞

x2 − 1 x + 2 = ? This time: lim

x→∞

x2 − 1 x + 2 = lim

x→∞

x2 − 1 x + 2 · x−1 x−1 = lim

x→∞

x(1 − x−1) 1 + 2x−1

?

= ( lim

x→∞ x)

• lim

x→∞

1 − x−1 1 + 2x−1

• ?

= ∞ · 1 ? = ∞

Limits at infinity of rational functions

lim

x→∞

x2 − 1 x + 2 = ? This time: lim

x→∞

x2 − 1 x + 2 = lim

x→∞

x2 − 1 x + 2 · x−1 x−1 = lim

x→∞

x(1 − x−1) 1 + 2x−1 = ( lim

x→∞ x)

• lim

x→∞

1 − x−1 1 + 2x−1

• = ∞ · 1 = ∞

Square-root tricks

Square-root tricks

lim

x→∞

• x2 − 2x + 3 − x = ?

Square-root tricks

lim

x→∞

• x2 − 2x + 3 − x = ?

= lim

x→∞

• x2 − 2x + 3 − x
• ·

√ x2 − 2x + 3 + x √ x2 − 2x + 3 + x

Square-root tricks

lim

x→∞

• x2 − 2x + 3 − x = ?

= lim

x→∞

• x2 − 2x + 3 − x
• ·

√ x2 − 2x + 3 + x √ x2 − 2x + 3 + x

• = lim

x→∞

x2 − 2x + 3 − x2 √ x2 − 2x + 3 + x = lim

x→∞

−2x + 3 √ x2 − 2x + 3 + x

Square-root tricks

lim

x→∞

• x2 − 2x + 3 − x = ?

= lim

x→∞

• x2 − 2x + 3 − x
• ·

√ x2 − 2x + 3 + x √ x2 − 2x + 3 + x

• = lim

x→∞

x2 − 2x + 3 − x2 √ x2 − 2x + 3 + x = lim

x→∞

−2x + 3 √ x2 − 2x + 3 + x = lim

x→∞

−2x + 3 √ x2 − 2x + 3 + x · x−1 x−1 = lim

x→∞

−2 + 3x−1 √ 1 − 2x−1 + 3x−2 + 1

Square-root tricks

lim

x→∞

• x2 − 2x + 3 − x = ?

= lim

x→∞

• x2 − 2x + 3 − x
• ·

√ x2 − 2x + 3 + x √ x2 − 2x + 3 + x

• = lim

x→∞

x2 − 2x + 3 − x2 √ x2 − 2x + 3 + x = lim

x→∞

−2x + 3 √ x2 − 2x + 3 + x = lim

x→∞

−2x + 3 √ x2 − 2x + 3 + x · x−1 x−1 = lim

x→∞

−2 + 3x−1 √ 1 − 2x−1 + 3x−2 + 1 = −2 √ 1 + 1 = −1

Square-root tricks

lim

x→∞

• x2 − 2x + 3 − x = −1