[height=2.3cm,width=2.3cm]IUGlogotrans Calculus A - One ...site.iugaza.edu.ps/biqelan/files/2019/09/MathA1301_sec2.4.pdf · Dr. Bisher M. Iqelan (IUG) Sec2.4: One-Sided Limits 1st

Calculus AOne-Sided Limits

Dr. Bisher M. [email protected]

Department of MathematicsThe Islamic University of Gaza

2019-2020, Semester 1

Dr. Bisher M. Iqelan (IUG) Sec2.4: One-Sided Limits 1st Semester, 2019-2020 1 / 12

One-Sided Limits

Sometimes saying that the limit of a function fails to exist at a point does not provideus with enough information about the behavior of the function at that particular point.

To see this, let us revisit the function g (x) =|x − 2|(x − 2)

introduced in the previous section

(Sec2.2). As we choose values of x close to 2, g (x) does not approach a single value,so the limit as x approaches 2 does not exist-that is, lim

x→2g (x) DNE. However, this

statement alone does not give us a complete picture of the behavior of the functionaround the x−value 2. To provide a more accurate description, we introduce the idea ofa one-sided limit. For all values to the left of 2 (or the negative side of 2), g (x) = −1.Thus, as x approaches 2 from the left, g (x) approaches −1.

Mathematically, we say that the limit as xapproaches 2 from the left is -1. Symbolically, weexpress this idea as

limx→2−

g (x) = −1.

Similarly, as x approaches 2 from the right (orfrom the positive side), g (x) approaches 1.Symbolically, we express this idea as

limx→2+

g (x) = 1.

We can now present an informal de�nition of one-sided limits.

Fact


One-Sided Limits

We de�ne two types of one-sided limits.Limit from the left:Let f (x) be a function de�ned at all values in an open interval (a, c), where a < c,and let L be a real number. If the values of the function f (x) approach the realnumber L as the values of x (where x < c) approach the number c, then wesay that L is the limit of f (x) as x approaches c from the left. Symbolically, weexpress this idea as lim

x→c−f (x) = L.

Limit from the right:Let f (x) be a function de�ned at all values in an open interval (c, b), where c < b,and let M be a real number. If the values of the function f (x) approach the realnumber M as the values of x (where x > c) approach the number c, then we saythat M is the limit of f (x) as x approaches c from the right. Symbolically, weexpress this idea as lim

x→c+f (x) = M.

Informal De�nition of One-Sided Limit

♣


Evaluating One-Sided Limits

For the function f (x) =

{x + 1 if x < 2

x2 − 4 if x ≥ 2, evaluate each of the following limits.

(a) limx→2−

f (x) (b) limx→2+

f (x)

Solution: We can use tables of functional values again Table. Observe that forvalues of x less than 2, we use f (x) = x + 1 and for values of x greater than 2, weuse f (x) = x2 − 4.

x f (x) = x + 1 x f (x) = x2 − 41.9 2.9 2.1 0.411.99 2.99 2.01 0.04011.999 2.999 2.001 0.0040011.9999 2.9999 2.0001 0.000400011.99999 2.99999 2.00001 0.0000400001

Based on this table, we can conclude that (a) limx→2− f (x) = 3 and (b)limx→2+ f (x) = 0. Therefore, the (two-sided) limit of f (x) does not exist at x = 2.The Figure below shows a graph of f (x) and reinforces our conclusion about theselimits.

Example1


Evaluating One-Sided Limits

Let f (x) =

x3 − 27

x − 3, if x < 3;

27, if x = 3;x2 + 21x − 27

x − 3, if x > 3.

Find limx→3

f (x).

Solution: Since f (x) is de�ned by di�erent expressions for x < 3 and x > 3, wehave to �nd limx→3+ f (x), and limx→3− f (x).

limx→3+ f (x) = limx→3+x2 + 21x − 27

x − 3

x → 3+ ⇒ x > 3 direct

substitution gives(0

0

)= limx→3+

(x − 3)(x + 24)

x − 3Factoring (x − 3)

= limx→3+��(x − 3)(x + 24)

��x − 3=27

limx→3− f (x) = limx→3−x3 − 27

x − 3

x → 3− ⇒ x < 3 direct

substitution gives(0

0

)= limx→3−

(x − 3)(x2 + 3x + 9

x − 3Factoring (x − 3)

= limx→3−��(x − 3)(x2 + 3x + 9

��x − 3=27

Hence limx→3 f (x) = 27 Since limx→3+ f (x) = 27 = limx→3− f (x)

Example1


Relating One-Sided and Two-Sided Limits

Let f (x) be a function de�ned at all values in an open interval containing c, withthe possible exception of c itself, and let L be a real number. Then,

limx→c

f (x) = L⇐⇒ limx→c−

f (x) = L and limx→c+

f (x) = L

Theorem

♣

Example2


Limit Does Not Exist from Either Side

Show that limx→0

sin

(1

x

)has no limit as x approaches 0 from either side. The Table

below lists values for the function sin(1x

)for the given values of x .

x sin(1x

)x sin

(1x

)-0.1 0.544021110889 0.1 -0.544021110889-0.01 0.50636564111 0.01 -0.50636564111-0.001 -0.8268795405312 0.001 0.8268795405312-0.0001 0.305614388888 0.0001 -0.305614388888-0.00001 -0.035748797987 0.00001 0.035748797987-0.000001 0.349993504187 0.000001 -0.349993504187

We can see that the y−values do not seem to approach any one single value, hence thelimit does not exist. Before drawing this conclusion, let's take the following sequence ofx−values approaching 0:

2

π,2

3π,2

5π,2

7π,2

9π,

2

11π, . . . .

The corresponding y-values are1,−1, 1,−1, 1,−1, . . . . Hence limx→0 sin

(1x

)does

not exist. The graph of f (x) = sin(1x

)is shown in

the Figure below and we can see that sin(1x

)oscillates between -1 and 1 as x approaches 0.

Example3


Limits Involvingsin x

x

For x in radians,

limx→0

sin x

x= 1

Theorem

♣

Evaluate limx→0

sin x

xusing a table of functional values.

Solution:

We have calculated the values of f (x) = sin xx for the values of x listed in Table below:

x sin xx x sin x

x-0.1 0.998334166468 0.1 0.998334166468-0.01 0.999983333417 0.01 0.999983333417-0.001 0.999999833333 0.001 0.999999833333-0.0001 0.999999998333 0.0001 0.999999998333

As we read down each sin xx column, we see that the values in each column appear to be

approaching one. Thus, it is fairly reasonable to conclude that limx→0

sin x

x= 1.

Example1



x

Show that (a) limh→0

cos h− 1

h= 0 and (b) lim

x→0

sin 2x

5x=

2

5Solution:

(a) Using the half-angle formula cos h = 1− 2 sin2(h/2) , we calculate

limh→0

cos h− 1

h= lim

h→0− 2 sin2(h/2)

h

= − limθ→0

sin θ

θsin θ Let θ = h/2

= −(1)(0) = 0

(b) Equation (1) does not apply to the original fraction. We need a 2x in the denom-inator, not a 5 x . We produce it by multiplying numerator and denominator by 2/5:

limx→0

sin 2x

5x= lim

x→0

(2/5) · sin 2x(2/5) · 5x

=2

5limx→0

sin 2x

2x

=2

5(1) =

2

5

Example2



x

Evaluate the following limits: 1. limx→0

sin(3x)

x2. lim

x→0

sin3 x

x3Solution:

1. Direct substitution will give us I.F. type(00

). The idea is to multiply and divide by

3, which will make the denominator the same as the argument of the sine function;

limx→0

sin(3x)

x= lim

x→0

3 sin(3x)

3x

= 3 limx→0

sin(3x)

3xFactor 3 and use the fact lim

x→0

sin(3x)

3x= 1

= 3(1) = 3

2. Direct substitution will give us I.F. type(00

). Remember that sin3 x

x3=(sin xx

)3. Now,

limx→0

sin3 x

x3= lim

x→0

(sin x

x

)3

Use the power law

=

(limx→0

sin x

x

)3

= (1)3 = 1

Example2



x

Evaluate the following limits limx→0

tan x

xand lim

x→0

x

tan xSolution:

limx→0

tan x

x= lim

x→0

sin xcos xx

Use the fact tan x =sin x

cos x

= limx→0

sin x

x cos xUse the fact

ab

c=

a

bc

= limx→0

[sin x

x

1

cos x

]Use the product low

=

(limx→0

sin x

x

)(limx→0

1

cos x

)Use the fact lim

x→0

sin x

xand direct substitution.

= (1)

(1

cos 0

)= 1

limx→0

x

tan x= lim

x→0

1tan xx

=1

1= 1

Example3



x

Evaluate the following limit: limx→0

sin(3x)

sin(4x)Solution:

limx→0

sin(3x)

sin(4x)= lim

x→0

sin(3x)x

sin(4x)x

Use the factb

c=

baca

and divide by x

= limx→0

3 sin(3x)3x

4 sin(4x)4x

=3 limx→0

sin(3x)

3x

4 limx→0

sin(4x)

4x

Use the quotient law.

=3(1)

4(1)=

3

4.

Example4


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[height=2.3cm,width=2.3cm]IUGlogotrans Calculus A - One ...site.iugaza.edu.ps/biqelan/files/2019/09/MathA1301_sec2.4.pdf · Dr. Bisher M. Iqelan (IUG) Sec2.4: One-Sided Limits 1st