Limits

From last time...

Let y = f (t) be a function that gives the position at time t of anobject moving along the y -axis. Then

Ave vel[t1

, t2

] =f (t

2

)� f (t1

)

t

2

� t

1

Velocity(t) = limh!0

f (t + h)� f (t)

h

.

a a+h

f(a)

f(a+h)

h

We need to be able to take limits!

Limit of a Function – Definition

We say that a function f approaches the limit L as x approaches a,

written limx!a

f (x) = L,

if we can make f (x) as close to L as we want by taking x

su�ciently close to a.

a

L

x

f(x)

Δx

Δy

i.e. If you need �y to be smaller,you only need to make �x smaller(� means ”change”)

One-sided limits

Lr

Ll

a

f(a)

Right-handed limit: Lr

= limx!a

+

f (x)

if f (x) gets closer to L

r

as x gets closer to a from the right

Left-handed limit: L` = limx!a

�f (x)

if f (x) gets closer to L` as x gets closer to a from the left

Theorem

The limit of f as x ! a exists if and only if both the right-hand

and left-hand limits exist and have the same value, i.e.

limx!a

f (x) = L if and only if limx!a

�f (x) = L and lim

x!a

+

f (x) = L.

Examples

limx!2

x � 2

x + 3

= 0

limx!1

x

2 � 1

x � 1

= 2

limx!0

1

x

is undefined

-4 4

-8

-4

4

8

-4 4

-8

-4

4

8

-4 4

-8

-4

4

8

Theorem

If limx!a

f (x) = A and limx!a

g(x) = B both exist, then

1. limx!a

(f (x) + g(x)) = limx!a

f (x) + limx!a

g(x) = A+ B

2. limx!a

(f (x)� g(x)) = limx!a

f (x)� limx!a

g(x) = A� B

3. limx!a

(f (x)g(x)) = limx!a

f (x) · limx!a

g(x) = A · B4. lim

x!a

(f (x)/g(x)) = limx!a

f (x)/ limx!a

g(x) = A/B (B 6=0).

In short: to take a limit

Step 1: Can you just plug in? If so, do it.

Step 2: If not, is there some sort of algebraic manipulation (likecancellation) that can be done to fix the problem? If so, do it.Then plug in.

Step 3: Learn some special limit to fix common problems. (Later)

If in doubt, graph it!

Examples

1. limx!2

x � 2

x + 3= 0 because if f (x) = x�2

x+3

, then f (2) = 0.

2. limx!1

x

2 � 1

x � 1!0

!0

If f (x) = x

2�1

x�1

, then f (x) is undefined at x = 1.However, so long as x 6= 1,

f (x) =x

2 � 1

x � 1=

(x + 1)(x � 1)

x � 1= x + 1.

So

limx!�1

x

2 � 1

x � 1= lim

x!1

x + 1 = 1 + 1 = 2 .

3. limx!0

px + 2�

p2

x

!0

!0

, so again, f (x) is undefined at a.

Multiply top and bottom by the conjugate:

limx!0

px + 2�

p2

x

= limx!0

px + 2�

p2

x

! px + 2 +

p2

px + 2 +

p2

!

= limx!0

x + 2� 2

x(px + 2 +

p2)

since (a � b)(a + b) = a

2 � b

2

= limx!0

x

x(px + 2 +

p2)

= limx!0

1px + 2 +

p2= 1

2

p2

Examples

3. limx!0

px + 2�

p2

x

!0

!0

, so again, f (x) is undefined at a.

Multiply top and bottom by the conjugate:

limx!0

px + 2�

p2

x

= limx!0

px + 2�

p2

x

! px + 2 +

p2

px + 2 +

p2

!

= limx!0

x + 2� 2

x(px + 2 +

p2)

since (a � b)(a + b) = a

2 � b

2

= limx!0

x

x(px + 2 +

p2)

= limx!0

1px + 2 +

p2= 1

2

p2

Examples

1. limx!1

x

2�3x+2

x

2

+4x�5

2. limx!�2

|x |x

= �1

3. limx!0

|x |x

is undefined

4. limx!0

(3 + x)2 � 32

x

= �1

6

= 6

Infinite limitsIf f (x) gets arbitrarily large as x ! a, then it doesn’t have a limit.Sometimes, though, it’s more useful to give more information.

Example:

For both f (x) = 1

x

and f (x) = 1

x

2

, limx!0

f (x) does not exist.However, they’re both “better behaved” than that might imply:

limx!0

+

1

x

= 1, limx!0

�

1

x

= �1 limx!0

+

1

x

2

= 1, limx!0

�

1

x

2

= 1

limx!0

1

x

does not exist limx!0

1

x

2

= 1

Why? A vertical asymptote occurs where

limx!a

+

f (x) = ±1 and limx!a

�f (x) = ±1

Infinite limitsBadly behaved example:

f (x) = csc(1/x)

-0.5 0.5

Zoom way in:

-0.005 0.005

(denser and denser vertical asymptotes)

limx!0

+

csc(1/x) does not exist, and limx!0

�csc(1/x) does not exist

Limits at InfinityWe say that a function f approaches the limit L as x gets biggerand bigger (in the positive or negative direction),

written limx!1

f (x) = L or limx!�1

f (x) = L

if we can make f (x) as close to L as we want by taking x

su�ciently large. (By “large”, we mean large in magnitude)

Example 1:

limx!�1

1

x

= 0 and limx!1

1

x

= 0.

Limits at Infinity: functions and their inverses

Theorem

If limx!a

± f (x) = 1, then limx!1 f

�1(x) = a.

If limx!a

± f (x) = �1, then limx!�1 f

�1(x) = a.

Example: Let arctan(x) be the inverse function to tan(x):y = tan(x):

π/2-π/2

Since limx!⇡/2� = 1

and limx!�⇡/2+ = �1

(restrict the domain to (�⇡2

, ⇡2

) to that we can invert)

y = arctan(x):

-π/2

π/2

we have limx!1 = ⇡/2

and limx!�1 = �⇡/2

Rational functionsLimits that look like they’re going to 1

1 can actually be doing lotsof di↵erent things. To fix this, divide the top and bottom by thehighest power in the denominator!

Ex 1 limx!1

x � 1

x

3 + 2!1!1

Fix: multiply the expression by 1/x3

1/x3 :

limx!1

x � 1

x

3 + 2= lim

x!1

✓x � 1

x

3 + 2

◆✓1/x3

1/x3

◆= lim

x!1

1

x

2

� 1

x

3

1 + 2

x

3

=0� 0

1 + 0= 0

Ex 2 limx!1

3x2 � 2x + 1

4x2 � 1!1!1

Fix: multiply the expression by 1/x2

1/x2

Ex 3 limx!1

x

4 � x

2 + 2

x

3 + 3!1!1

Fix: multiply the expression by 1/x3

1/x3

Rational functions: quick trick

limx!1

x � 1

x

3 + 2= 0 lim

x!1

3x2 � 2x + 1

4x2 � 1=

3

4lim

x!1

x

4 � x

2 + 2

x

3 + 3= 1

Suppose

P(x) = a

n

x

n+· · ·+a

1

x+a

0

and Q(x) = b

m

x

m+· · ·+b

1

x+b

0

are polynomials of degree n and m respectively. Then in general,

limx!1

P(x)

Q(x)= lim

x!1

a

n

x

n

b

m

x

m

= limx!1

a

n

b

m

x

n�m =

8><

>:

0 n < m

a

n

b

m

n = m

±1 n < m

Examples: Other ratios with powers.Example:

limx!1

xp3x2 + 2

[hint: multiply by 1/x1/x and remember a

pb =

pa

2

b.]

limx!1

xp3x2 + 2

= limx!1

✓xp

3x2 + 2

◆✓1/x

1/x

◆

= limx!1

11

x

p3x2 + 2

= limx!1

1q1

x

2

(3x2 + 2)

= limx!1

1q3 + 2

x

2

=1p3 + 0

= 1p3

Evaluating limits when x ! 0.

1. Show lim

x!0(x

2 � 2)

2+ 6 = 10.

2. Show lim

x!0

5x

x

= 5.

3. Show lim

x!0

17x

2x

= 17/2.

4. Show lim

x!0

�317x

422x

= �317/422.

5. Show lim

x!0

�317x� 3

422x+ 5

= �3/5.

6. Show lim

h!0

px+ h�

px

h

=

1

2

px

.

7. Show lim

x!0

p1 + x+ x

2 � 1

x

= 1/2.

8. Show lim

x!0

p3 + x�

p3

x

= 1/(2

p3).

9. Show lim

h!0

1

h

✓1p

x+ h

� 1px

◆= �(1/2)x

�3/2.

10. Show lim

x!0

2xpa+ x�

pa� x

= 2

pa.

11. Show lim

x!0

p1 + x� 1

x

= 1/2.

12. Show lim

x!0

xp1 + x� 1

= 2.

13. Show lim

x!0

e

x

+ e

�x � 2

x

2= 1.

14. Show

lim

h!0

f(x+ h)� f(x)

h

=

a

2

pax+ b

when f(x) =

pax+ b.

15. Show

lim

h!0

f(x+ h)� f(x)

h

= mn(mx+ c)

n�1

when f(x) = (mx+ c)

n

.

Evaluating limits when x! a.

1. Show lim

x!1(6x

2 � 4x + 3) = 5.

2. Show lim

x!7

x

2 � 49

x� 7

= 14.

3. Show lim

x!2

x

2 � 6x + 8

x� 2

= �2.

4. Show lim

x!�5

2x

2+ 9x� 5

x + 5

= �11.

5. Show lim

x!1

x

3 � 1

x� 1

= 3.

6. Show lim

x!3

x

2 � 4x + 3

x

2 � 2x� 3

= 1/2.

7. Show lim

x!�2

x

3+ 8

x + 2

= 4.

8. Show lim

x!3

x

4 � 81

x� 3

= 108.

9. Show lim

x!5

x

5 � 3125

x� 5

= 3125.

10. Show lim

x!a

x

12 � a

12

x� a

= 12a

11.

11. Show lim

x!a

x

5/2 � a

5/2

x� a

= (5/2)a

3/2.

12. Show lim

x!a

(x + 2)

5/3 � (a + 2)

5/3

x� a

= (5/3)(a + 2)

2/3.

13. Show lim

x!4

x

3 � 64

x

2 � 16

= 6.

14. Show lim

x!2

x

5 � 32

x

3 � 8

= 20/3.

15. Show lim

x!1

x

n � 1

x� 1

= n.

16. Show lim

x!a

px�

pa

x� a

=

1

2

pa

.

17. Show lim

x!2

p3� x� 1

2� x

= 1/2.

18. Show lim

x!a

pa + 2x�

p3xp

3a + x� 2

px

=

2

p3

9

.

19. Show lim

x!a

x

n � a

n

x� a

= na

n�1.

Evaluating limits when x!1.

1. Show lim

x!1

x + 2

x� 2

= 1.

2. Show lim

x!1

3x

2+ 2x� 5

5x

2+ 3x + 1

= 3/5.

3. Show lim

x!1

x

2 � 7x + 11

3x

2+ 10

= 1/3.

4. Show lim

x!1

2x

3 � 5x + 7

7x

3+ 2x

2 � 6

= 2/7.

5. Show lim

x!1

(3x� 1)(4x� 5)

(x + 6)(x� 3)

= 12.

6. Show lim

x!1

xp4x

2+ 1� 1

= 1/2.

7. Show lim

x!�12

x

= 0.

8. Show lim

t!1

t + 1

t

2+ 1

= 0.

9. Show lim

n!1

pn

2+ 1� n = 0.

10. Show lim

n!1

pn

2+ n� n = 1/2.