Limits and ContinuityMath 131

Limits and Continuity – p. 1/23

Average velocity

To explore the concept of limits, we look atphenomenon like average velocity, where limitsoccur naturally.

Suppose we were to find the average velocity of arock free falling from rest between the timet1 andt2second.

From Newton’s second law, the distance travelled bythe rock in timet second is

x(t) =1

2gt2 = 4.9t2

Limits and Continuity – p. 2/23

Average velocity...

The average velocity betweent1 andt2 is given by:

∆x

∆t=

x(t2) − x(t1)

t2 − t1=

4.9(t22− t2

1)

t2 − t1

In the first 2 seconds (time interval[0, 2]) theaverage velocity is:

∆x

∆t=

4.9(22) − 4.9(02)

2 − 0= 9.8 m/s

Limits and Continuity – p. 3/23

Instantaneous Velocity

To find the instantaneous velocity at timet = 1, wewrite the average velocity in the time interval[1, 1 + h].

∆y

∆t=

4.9(1 + h)2 − 4.9(1)2

h

Now we leth decrease slowly to0 and see if theaverage velocity get closer to a particular value

h 1 0.1 0.01 0.001 0.0001∆y/∆t 14.7 10.29 9.8490 9.8049 9.8005

This suggest that at timet = 1 the rock is falling at9.8 m/s Limits and Continuity – p. 4/23

Instantaneous Velocity . . .

Similarly we can find the instantaneous velocity att = 2 by looking at the average velocities in theinterval[2, 2 + h] ash tends to0.

∆y

∆t=

4.9(2 + h)2 − 4.9(2)2

h

h 1 0.1 0.01 0.001 0.0001∆y/∆t 24.5 20.09 19.649 19.6049 19.6005

This suggest that at timet = 2 the rock is falling at19.6 m/s

Limits and Continuity – p. 5/23

Instantaneous Velocity . . .

The velocity at timet = 0 is limit of the averagevelocity ash approaches zero.

Velocity at time t = limh→0

4.9(t + h)2 − 4.9t2

h

We can’t just find the limit by replacingh by 0(avoid the division by 0), we first have to do somealgebraic manipulations

4.9(t + h)2 − 4.9t2

h=

4.9(t2 + 2th + h2 − t

2)

h=

4.9(2th + h2)

h= 9.8t + 4.9h

In particular whent = 1 the velocity is 9.8(9.8 × 1 + 4.8 × 0)

Limits and Continuity – p. 6/23

Informal definition of limits

DefinitionIf f(x) is defined for allx neara except possibly ata itself,

and iff(x) is close enough toL asx get close toa, but not

equal toa, we say that the functionf approachesL asx

approachesa, and we write

limx→a

f(x) = L.

Example

limx→a

x2 = a2; limx→a

(x + 1) = a + 1

This shows that the limit can sometimes beevaluated by just takingf(a).

Limits and Continuity – p. 7/23

Informal definition of limits . . .

In situations whenf(x) is undefined, algebraic manipulations

are used

Example

limx→3

x2 − x − 6

x − 3= lim

x→3

(x + 2)(x − 3)

x − 3= lim

x→3(x + 2)

= 6

Limits and Continuity – p. 8/23

Informal definition of limits . . .

limx→4

√x − 2

x2 − 16= lim

x→4

(√

x − 2)(√

2 + x)

(x2 − 16)(√

2 + x)

= limx→4

(√

x − 2)(√

2 + x)

(x2 − 16)(√

2 + x)

= limx→4

x − 4

(x − 4) (x + 4)(√

x + 2)

= limx→4

1

(x + 4)(√

x + 2)

=1

(4 + 4)(√

4 + 2)

=1

32Limits and Continuity – p. 9/23

Informal definition of limits . . .

Even iff is defined atx = a, the limit of f(x) asx approaches

a may not be equal tof(a)

Example

Let

g(x) =

x ; x 6= 2

1 ; x = 2

limx→2

g(x) = limx→2

x = 2 although g(2) = 1.

limx→0

1

xdoes not exist(x = 0 is a vertical asymptote)

Limits and Continuity – p. 10/23

Left and right limit

If f(x) is defined on some interval(b, a) extending to the left

of x = a, and iff(x) is as close toL by takingx to the left ofa

and close enough toa, then we sayf(x) hasleft limit atx = a,

and we write

limx→a−

f(x) = L.

If f(x) is defined on some interval(a, b) extending to the right

of x = a, and iff(x) is as close toL by takingx to the right of

a and close enough toa, then we sayf(x) hasright limit at

x = a, and we write

limx→a+

f(x) = L.

Limits and Continuity – p. 11/23

Left and Right limits

Example

The Heaviside function

H(x) =

1 ; x ≥ 0

0 ; x < 0

Has left limit0 and right limit1.

limx→0−

H(x) = 0 and limx→0+

H(x) = 1

Since the left and right limits ofH(x) are not equal asx

approaches0, then

limx→0

H(x) does not exist.

Limits and Continuity – p. 12/23

Left and Right limits

Theorem

A functionf(x) has limitL atx = a if and only if the left and

right limits exist and are both equal toL.

limx→a

f(x) = L ⇐⇒ limx→a−

f(x) = limx→a+

f(x) = L.

Example

If f(x) = |x|x

,

limx→0−

f(x) = limx→0

−x

x= −1 and lim

x→0+f(x) = lim

x→0

x

x= 1

Since

limx→0−

f(x) 6= limx→0+

f(x), thenlimx→0

f(x) does not exist

.Limits and Continuity – p. 13/23

Consider the functionf(x) =√

x. This function is defined for

x > 0, so it cannot have a limit atx = 0 (left limit does not

exist), though it has a right limit atx = 0.

Limits and Continuity – p. 14/23

Properties of limits

If

limx→a

f(x) = L , limx→a

g(x) = M and k is a constant

Limit of a sum: limx→a

(f(x) + g(x)) = L + M

Limit of a difference: limx→a

(f(x) − g(x)) = L − M

Limit of a product: limx→a

(f(x)g(x)) = LM

Limit of a multiple: limx→a

kf(x) = kL

Limit of a quotient: limx→a

f(x)

g(x)=

L

M, if M 6= 0

If f(x) ≤ g(x) : L ≤ M (order is preserved).

Limits and Continuity – p. 15/23

The squeeze theorem

Theorem

Suppose thatg(x) ≤ f(x) ≤ h(x) holds for allx in some open

interval containinga, except possibly atx = a. Suppose also

that

limx→a

g(x) = limx→a

h(x) = L.

Then

limx→a

f(x) = L

Example:Given3 − x2 ≤ u(x) ≤ 3 + x2, Since

limx→0

(3 − x2) = limx→0

(3 + x2) = 3,

the squeeze theorem implies thatlimx→0

u(x) = 3.Limits and Continuity – p. 16/23

Limit at infinity and infinite limits

Consider the functionf(x) = x√x2+1

, let observe the values off(x)

asx becomes large, positively or negatively.

x -1000 -100 -10 -1 0

f(x) - 0.9999995 -0.9999500 -0.9950372 -0.7071068 0

limx→−∞

f(x) = −1

x 0 1 10 100 1000

f(x) 0 0.7071068 -0.9950372 0.9999500 0.9999995

limx→∞

f(x) = 1

Limits and Continuity – p. 17/23

Limit at infinity and infinite limits

If the functionf is defined on(a, ∞) and iff(x) is as close as

we want toL by takingx large enough, then we say thatf(x)

approaches the limitL as x approaches infinityand we write

limx→∞

f(x) = L.

If the functionf is defined on(−∞, a) and iff(x) is as close

as we want toM by takingx negative and large enough in

absolute value, then we say thatf(x) approaches the limitMas x approaches negative infinityand we write

limx→−∞

f(x) = M.

Example:limx→∞1x

= limx→−∞1x

= 0 (y = 0 is an horizontal

asymptote) Limits and Continuity – p. 18/23

Limits at infinity for rational frac-tions

Numerator and denominator of same degree

limx→±∞

2x2 − x + 3

3x2 + 5= lim

x→±∞

x2(2 − (1/x) − (3/x2))

x2(3 + (5/x2))=

2

3

Degree of numerator less than degree of denominator

limx→±∞

5x + 2

2x3 − 1= lim

x→±∞

x3 ((5/x2) + (2/x3))

x3 (2 − (1/x3))= 0

Limits and Continuity – p. 19/23

Infinite limits

A function whose value grows arbitrarily large is said to have an

infinite limit. Consider the functionf(x) = 1x2 asx approaches 0

from either sides.

x -0.1 -0.01 -0.001 -0.0001

f(x) 100 10000 1,000,000 100,000,000

x 0.1 0.01 0.001 0.0001

f(x) 100 10000 1,000,000 100,000,000This suggest that

limx→0

1

x2= ∞

Limits and Continuity – p. 20/23

Infinite limits

Consider the functionf(x) = 1

x

As x approaches 0 from the rightf(x) becomeslarger and larger positively:

limx→0+

f(x) = ∞.

As x approaches 0 from the leftf(x) becomes largerand larger negatively:

limx→0−

f(x) = −∞.

Limits and Continuity – p. 21/23

Polynomial behaviour at infinity

The highest degree term of the polynomial determines the limit of

the whole polynomial at∞ and−∞.

Example:

limx→∞

(3x3−x2+2) = limx→∞

3x3

(

1 − 1

3x+

2

3x2

)

= limx→∞

3x3 = ∞

limx→−∞

(3x3 − x2 + 2) = limx→−∞

3x3

(

1 − 1

3x+

2

3x2

)

= −∞

limx→±∞

(x4 − 5x3 + x) = limx→±∞

x4 = ∞

Limits and Continuity – p. 22/23

Rational fraction with numerator ofhigher degree

We divide the numerator and the denominator by the largest power

of x in the denominator:

limx→∞

x3 + 1

x2 + 1= lim

x→∞

x + (1/x2)

1 + (1/x2)= ∞

Limits and Continuity – p. 23/23