Unit I Review
Mathematics 21
Institute of Mathematics (UP Diliman)
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 1 / 20
Review concepts and practice solving problems involving:
1 Limit of a Function (Algebraic, Piece-wise, Circular, Inverse Circular,
Hyperbolic and Inverse Hyperbolic) at a Point
2 One-Sided Limits
3 Infinite Limits
4 Limits at Infinity
5 Continuity of a Function at a Point
6 Continuity of a Function in an Interval
7 Intermediate Value Theorem
8 Squeeze Theorem
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 2 / 20
Reminder
First Long Exam: 06 September 2018 (tomorrow)
Discussion class schedule Exam schedule
7:15 - 8:15 7:00-8:20
8:45 - 9:45 8:30-9:50
10:15 - 11:15 10:00-11:20
11:45 - 12:45 11:30-12:50
1:15 - 2:15 1:00-2:20
2:45 - 3:45 2:30-3:50
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 3 / 20
More reminders
UP blue book for solutions, answers, and scratch work
No loose sheets (do not tear off any page of the blue book)
Black or blue pens only
No early submission
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 4 / 20
Formulas and graphs to be provided
Graphs of hyperbolic and inverse hyperbolic functions
Inverse hyperbolic functions in terms of natural logarithms
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 5 / 20
3x2 − x − 10
Evaluate lim
x →2− | x − 2|
Solution: Note that | x − 2| = x − 2 when x ≥ 2
and | x − 2| = −( x − 2) when x ≤ 2
3x2 − x − 10 3x2 − x − 10 ( 00 )
lim = lim
x →2− | x − 2| x →2− −( x − 2)
(3x + 5)( x − 2)
= lim
x →2− −( x − 2)
= lim −(3x + 5) = −11
x →2−
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 6 / 20
4 2x + 8
Evaluate lim −
x →−2+ x + 2 x2 + 5x + 6
Solution:
4 2x + 8 4 2( x + 4) (∞−∞)
lim − 2 = lim −
x →−2+ x + 2 x + 5x + 6 x →−2+ x + 2 ( x + 2)( x + 3)
4( x + 3) − 2( x + 4)
= lim
x →−2+ ( x + 2)( x + 3)
2x + 4 ( 00 )
= lim
x →−2+ ( x + 2)( x + 3)
2
= lim
x →−2+ x+3
= 2
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 7 / 20
√
3− 4x2 − 5x
Evaluate lim
x →−∞ 6x − 2
Solution:
√ √
√1 3
− √1 4x2 − 5x
3− 4x2 − 5x x2 −x x2
lim · = lim
x →−∞ 6x − 2 √1 x →−∞ −6 − −2x
x2
q
3
−x − 4 − 5x
= lim 2
x →−∞ −6 + x
√
0− 4−0 1
= =
−6 + 0 3
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 8 / 20
1 − cos(3x − 6)
Evaluate lim
x →2 3x2 − 5x − 2
Solution:
1 − cos(3x − 6) 1 − cos(3x − 6) 3
lim = lim ·
x →2 3x2 − 5x − 2 x →2 ( x − 2)(3x + 1) 3
1 − cos(3x − 6)
1
= 3 lim
x →2 (3x − 6) (3x + 1)
= 3 (0) (1/7) = 0
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 9 / 20
Discuss the continuity of the following function at its possible points discontinuity.
Classify each discontinuity and redefine the function in case of a removal
discontinuity.
1 1
+ , x<0
x2 −x x
f (x) = [[ x − 1]] , 0≤x≤2
√
x − 2, x>2
Note:
−1, 0 ≤ x < 1
[[ x − 1]] = 0, 1≤x<2
1, x=2
Possible points of discontinuity: x = 0, x = 2, and x = 1.
f is sure to be continuous over R − {0, 1, 2}.
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 10 / 20
1 1
+ , x<0
x2 −x x
f (x) = [[ x − 1]] , 0≤x≤2
√
x − 2, x>2
At x = 0:
1 f (0) = [[0 − 1]] = −1
1+x−1
1 1 1
2 lim f ( x ) = lim + = lim = lim = −1
x →0− x →0− x 2 − x x x →0− x ( x − 1 ) x →0− x − 1
3 lim f ( x ) = lim [[ x − 1]] = lim (−1) = −1
x →0+ x →0+ x →0+
Therefore, f is continuous at x = 0.
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 11 / 20
1 1
+ , x<0
x2 −x x
f (x) = [[ x − 1]] , 0≤x≤2
√
x − 2, x>2
At x = 2
1 f (2) = [[2 − 1]] = 1
2 lim f ( x ) = lim [[ x − 1]] = lim 0 = 0
x → 2− − x →2 − x →2
√ √
3 lim f ( x ) = lim x−2 = 0 ( 0+ )
x → 2+ x → 2+
Therefore, f has a removable discontinuity at x = 2.
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 12 / 20
1 1
+ , x<0
x2 − x
x
f (x) = [[ x − 1]] , 0≤x≤2
√
x − 2, x>2
At x = 1
1 f (1) = [[1 − 1]] = 0
2 lim f ( x ) = lim [[ x − 1]] = lim −1 = −1
x → 1− − x →1 − x →1
3 lim f ( x ) = lim [[ x − 1]] = lim 0 = 0
x → 1+ + x →1 + x →1
Thus, f has a jump essential discontinuity at x = 1.
(But f is continuous from the right at x = 1.)
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 13 / 20
x2 cos(2x2 + 1)
Use the Squeeze Theorem to evaluate lim .
x →+∞ 2x4 + 1
Solution:
−1 ≤ cos(2x2 + 1) ≤ 1
− x2 ≤ x2 cos(2x2 + 1) ≤ x2
− x2 x2 cos(2x2 + 1) x2
≤ ≤
2x4 + 1 2x4 + 1 2x4 + 1
− x2 x2
lim =0 and lim =0
x →+∞ 2x4 + 1 x →+∞ 2x4 +1
x2 cos(2x2 + 1)
Thus, lim = 0.
x →+∞ 2x4 + 1
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 14 / 20
Using the Intermediate Value Theorem, show that the equation
x
x2 − cos( x2 − 3x ) + =0
x+2
has a solution between x = 0 and x = 3.
Solution:
x
1 Is the function f ( x ) = x2 − cos( x2 − 3x ) + continuous on [0, 3]?
x+2
Answer: YES, since f is defined over the interval [0, 3]; f will only be
undefined when x = −2 which is not in [0, 3].
2 Is f (0) 6= f (3)?
0
Answer: YES, since f (0) = 02 − cos 0 + = −1 while
0+2
3 43
f (3) = 32 − cos 0 + = .
5 5
43
Since 0 is between f (0) = −1 and f (3) = , then by the IVT, there exists a
5
c ∈ (0, 3) such that f ( x ) = 0.
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 15 / 20
True or False?
1 If the functions f and g are continuous at x = a, then f ◦ g is continuous at
x = a.
1
Answer: FALSE . Suppose f ( x ) = and g( x ) = x − 2.
x
1
Then ( f ◦ g)( x ) = which is not continuous at x = 2.
x−2
sin x
2 If f ( x ) = , then lim f ( x ) does not exist.
|x| x →0
sin x
Answer: TRUE. lim f ( x ) = lim =1
x →0+ x →0+ x
sin x
lim f ( x ) = lim = −1
x →0 − x →0 − −x
Since lim f ( x ) 6= lim f ( x ), then lim f ( x ) dne.
x →0+ x →1− x →0
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 16 / 20
Evaluate the following.
π √ 3π π
a. cos−1 1
2 = d. cos−1 − 22 = g. cos−1 (0) =
3 √ 4 2
−
1 1 π π
b. sin 2 = e. sin−1 − 22 = −
6 √ 4 h. tan−1 (0) = 0
π π
− 1
c. tan (1) = f. sin−1 − 23 = −
4 3
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 17 / 20
Evaluate the following limits.
1 lim (100 − x2 − ln x ) = +∞ 10 lim
x + [[2x ]]
= −∞
x →0+ | x + 1|
x →−1−
2e x − 1
2 lim =2 x− π
x →+∞ e x + 2 3
x x
11 limπ =1
5π
5 +3 x→ 3 cos −x
3 lim =0 6
x →+∞ 7x − 2x
1−x
1
4 lim cos −1
=
2π 12 lim x3 cos √ =0
x →1 x2 − 1 3 x →0+ x
x
1
p
5 lim tan−1 =
π 13 lim 2x + x2 + 2x = −∞
x →−∞ x →−∞
2 2
6 −1
lim cosh(coth x ) = 1 14 lim (e− tan x ) = 0
x →+∞ x → π2 −
7 lim sinh−1 (ln(1 − x )) = −∞
x →1− 15 lim cos 31 cot−1 1x =
8 lim [ln( x2 + x + 1) − ln( x + 1) − x →0−
x →+∞ π
ln( x + 2)] = 0 cos = 1/2
3
sin( x ) −π
9 lim p = −4 16 lim tan−1 (csch x ) =
x →0− |x| + 4 − 2 x →0 − 2
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 18 / 20
Try at home
Find and classify (as removable, jump essential or infinite essential) all points of
discontinuity of the function f if f is given by
2 + x − x2
x≤0
,
x+1
f (x) = [[2x ]] + 1 , 0<x≤1
9x2 − 18x + 9
, x>1
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 19 / 20
GOODLUCK!!!
(This slide can be found at http://sites.google.com/site/updmathsite)
Institute of Mathematics (UP Diliman) Unit I Review Mathematics 21 20 / 20
