M53-Unit1-Review.pdf

Unit I Review

Source: M53 1st Sem AY 2015-2016

Mathematics 53

Review concepts and practice skills:

Limit of a Function at a Point

One-Sided Limits

Infinite Limits

Limits at Infinity

Continuity of a Function at a Point

Continuity of a Function in an Interval

Intermediate Value Theorem

Squeeze Theorem

Evaluate lim x→2−

3x2−x−10

|x−2|

Solution: Note that|x−2|=x−2whenx≥2

and|x−2|=−(x−2)whenx≤2

lim x→2−

3x2−x−10

|x−2| = xlim→2−

3x2−x−10

−(x−2)

0− 0+

= lim

x→2−

(3x+5)(x−2)

−(x−2)

= lim

Evaluate lim x→−2+

4

x+2−

2x+8

x2_+5x+6

Solution:

lim x→−2+

4

x+2−

2x+8

x2_+5x+6

= lim

x→−2+

4

x+2−

2(x+4) (x+2)(x+3)

(∞−_∞)

= lim

x→−2+

4(x+3)−2(x+4) (x+2)(x+3)

= lim

x→−2+

2x+4

(x+2)(x+3)

(0 0)

= lim

x→−2+

2

x+3

Evaluate lim x→−∞

3−√4x2_−5x

6x−2

Solution:

lim x→−∞

3−√4x2_−5x

6x−2 ·

1

x

1

x

= lim

x→−∞

3

x−1x √

4x2_−5x

6− 2

x

= lim

x→−∞

3

x−

−_√1

x2

√

4x2_−5x

6−2_x

= lim

x→−∞

3

x+

q

4−5

x 6−2_x

= 0+

√ 4−0 6−0 =

Evaluatelim x→2

1−cos(3x−6)

3x2_−5x−2

Solution:

lim x→2

1−cos(3x−6)

3x2_−5x−2 = lim_x→2

1−cos(3x−6) (x−2)(3x+1) ·

3 3

= 3 lim x→2

1−cos(3x−6) (3x−6)

1

(3x+1)

Discuss the continuity of the following function at 3 possible points discontinuity.

Classify each discontinuity and redefine the function in case of a removal

discontinuity.

f(x) = 

         

         

x2−3x+2

2−x ,x<2

[[5−2x]]−x

2 , 2≤x<3

x2−4x+1

2 ,x≥3

Possible points of discontinuity: x=2, x= 5

f(x) =                     

x2−3x+2

2−x ,x <2

[[5−2x]]−x

2 , 2≤x<3

x2_−4x+1

2 ,x ≥3

Atx =2:

a) f(2) = [[5−2(2)]]−2 2 =0

b) lim

x→2− f(x) =_xlim_→2−

x2_−3x+2

2−x

= lim

x→2−

(x−1)(x−2)

2−x =−1

c) lim

x→2+ f(x) =_xlim_→2+[[5−2x]]− x

2

=−1 ([[1−]] =0)

f(x) =                     

x2−3x+2

2−x ,x <2

[[5−2x]]−x

2 , 2≤x<3

x2−4x+1

2 ,x ≥3

Atx = 5

2: a) f 5 2

=−5

4

b) lim x→5

2

− f(x) = lim

x→5 2

−[[5−2x]]− x

2

=−5

4 ([[0

+_{]] =0)}

c) lim x→5

2

+ f(x) = lim

x→5 2

+[[5−2x]]− x

2

=−9

4 ([[0

−_{]] =−1)}

f has a jump essential discontinuity atx= 5

f(x) =                     

x2−3x+2

2−x ,x <2

[[5−2x]]−x

2 , 2≤x<3

x2−4x+1

2 ,x ≥3

Atx =3:

a) f(3) =32−4(3) +1

2 =− 5 2

b) lim

x→3− f(x) =_xlim_→3−[[5−2x]]− x

2

=−5

2 ([[−1

+_{]] =−1)}

c) lim

x→3+ f(x) =_xlim_→3+x

2_−4x+1

2

=−5

2

f(x) = 

         

         

x2_−3x+2

2−x ,x <2

[[5−2x]]−x

2 , 2≤x<3

x2−4x+1

2 ,x ≥3

Summary:

a) f has a removable discontinuity at

x=2

b) f has a jump essential

discontinuity atx= 5₂

c) f is continuous atx=3

Redefining,

F(x) = (

f(x) ,x6=2

−1 ,x=2

Use the Squeeze Theorem to evaluate lim x→+∞

x2cos(2x2+1)

2x4₊₁

Solution:

−1 ≤ cos(2x2_{+1) ≤ 1}

−x2 _{≤ x}2_cos(2x2_{+1) ≤ x}2

−x2

2x4₊₁ ≤

x2_cos(2x2₊₁₎

2x4₊₁ ≤

x2

2x4₊₁

lim x→+∞

−x2

2x4₊₁ =0 and _x→lim+∞

−x2

2x4₊₁ =0

Thus, lim

x→+∞

x2cos(2x2+1)

Using the Intermediate Value Theorem, show that the equation

x2−cos(x2−3x) + x x+2 =0

has a solution betweenx =0andx =3.

Solution:

a) Is the function f(x) =x2_−cos(x2_{−3x) +} x

x+2 continuous on[0, 3]?

Answer: YES, since f is defined over the interval[0, 3]; f will only be

undefined whenx=−2which is not in[0, 3].

b) Is f(0)6= f(3)?

Answer: YES, since f(0) =02−cos 0+ 0

0+2 =−1while

f(3) =32−cos 0+3

5 = 43

5 .

True or False?

1 _{If the functions} f _andg_{are continuous at}x =a_{, then} f◦g_{is continuous at}

x=a.

Answer: FALSE . Suppose f(x) = 1

x andg(x) =x−2.

Then(f◦g)(x) = 1

x−2 which is not continuous atx=2.

2 _If f(x) = sinx

|x| , thenlimx→0f(x)does not exist.

Answer: TRUE. lim

x→0+ f(x) =_xlim_→0+

sinx

x =1

lim

x→0− f(x) =_xlim_→0−

sinx

−x =−1

2013 Sample Exam Questions

Evaluate the following limits.

1. lim x→−1−

x+ [[2x]]

|x+1|

2. lim x→π

3

x− π

3

cos 5π

6 −x

3. lim x→0+x

3_cos√1

x

4. lim x→−∞

2x+px2_+2x

Find and classify (as removable, jump essential or infinite essential) all points of

discontinuity of the function f if f is given by

f(x) = 

   

   

2+x−x2

x+1 , x≤0