Performance Task on Basic Calculus

Arganda, Asunan, Dagoy, Panlilio, Parocha, Secarro, Villacarlos

February 27, 2023

For this performance task, we are asked to create 5 illustrative examples similar to the limit problems
presented on the following videos.

1 Limit at a point of discontinuity — Khan Academy

(https://youtu.be/Y7sqB1e4RBI)
The video discussed the concept of discontinuity in an indirect manner. Instead of fully showcasing the
definition of what a limit at a point of discontinuity is, it gave an example of a discontinuous function,
and tried to find the limit at its point of discontinuity. Based on my observation, I think that the point of
discontinuity defines where the function would break, or it is the part of the Cartesian plane that isn’t part
of the domain of the function.

3 x2
lim− f (x) = 3 f (x) = lim− f (x) = 3 f (x) =
x→3
x−1 x→3 x2
−4
lim+ f (x) = −2 lim− f (x) = −∞ lim f (x) = −2 lim − f (x) = +∞
x→1 x→−2
x→3 x→3+
lim f (x) = DN E lim+ f (x) = +∞ lim f (x) = DN E lim f (x) = −∞
x→1 x→−2+
x→3 x→3
f (x) = 1 lim f (x) DN E f (3) = 1 lim f (x) DN E
x→1
x→−2

6x2 + 13x + 5
f (x) =
3x + 5
|2x − 4| 7
lim f (x) = 3 f (x) = lim f (x) = 3 lim − f (x) = −
x→3−
2x − 4 x→3− 5 3
x→− 3
lim f (x) = −2 lim f (x) = −1 lim f (x) = −2
x→2− 7
x→3+ x→3+ lim + f (x) = −
lim f (x) = 1 x→− 53 3
lim f (x) = DN E x→2+ lim f (x) = DN E
x→3 x→3
7
f (x) = 1 lim f (x) DN E f (3) = 1 lim 5 f (x) = −
x→2 x→− 3 3

f (x) = csc x
lim− f (x) = 3 lim f (x) = −∞
x→3 −
x→ 3π
2
lim f (x) = −2
x→3+ lim f (x) = +∞
+
x→ 3π
lim f (x) = DN E 2
x→3
lim f (x) DN E
f (x) = 1 x→ 3π
2

2 3-step continuity test — The Organic Chemistry Tutor

(https://youtu.be/WT7oxiiFYt8)
The functions shown in the video are piecewise functions. Evaluating them by using the 3-step continuity
test can help us determine whether the function is continuous or discontinuous. The first step is to determine
whether the function is defined at a given point. The second step is to show that the limit of f(x) exists, this
means that the limit from the left side and the right side should be equal. If the limits are different, it may
be a discontinuity, specifically, a jump discontinuity or an infinite discontinuity. For the third step, the limit
of f(x) should be equal to the function at a given point. However, if the limit does exist but is not equal to
the function at a given point, then it is a removable discontinuity.

1
 Is the function Determine whether the function
(
x3 − 2x + 1 x ≤ 2
(
f (x) = 4x − x2 x ≤ 0
3x − 2 x>2 f (x) =
4+x x>0
discontinuous as x = 2?
is continuous or discontinuous at x = 0.
1.
1.
f (2) = x3 − 2x + 1
= 23 − 2(2) + 1 f (0) = 4x − x2 = 0
=8−4+1=5
2.
2.
lim 4x − x2 = 0
lim x3 − 2x + 1 = 5 x→0−
x→2−
lim 4 + x = 4
lim+ 3x − 2 = 4 x→0+
x→2
lim f (x) = DN E
lim f (x) = DN E x→0
x→2

∴ Yes, the function is discontinuous at f (2). ∴ The function is discontinuous at x = 0.

Find out the type of discontinuity exhibited by the Find if the following function breaks at x = 3.
graph of

3
−x − 1 x < −2

2
2x + 5 x<3
 
f (x) = 3 x = −2 f (x) = 20 + x x=3

4x + 15 x > −2
 
 3
x −4 x>3
at x = −2.
1.
1.
f (3) = 20 + x = 23
f (−2) = 3

2. 2.

lim −x3 − 1 = −(−2)3 − 1 = 8 − 1 = 7 lim 2x2 + 5 = 2(3)2 + 5 = 23

x→−2− x→3−
lim 4x + 15 = 4(−2) + 15 = 7 lim x3 − 4 = 33 − 4 = 23
x→−2+ x→3+
lim f (x) = 7 lim f (x) = 23
x→−2 x→3

3. 3.
lim f (x) = 7 ̸= f (−2) = 3
x→−2 lim f (x) = f (3) = 23
x→3
The function fails the third test ∴ it exhibits a
removable discontinuity. ∴ The function doesn’t break at x = 3.
9
Show that f (x) = x−3 is discontinuous at x = 3.

Simply, the function is discontinuous at x = 3 because we cannot evaluate f (3), thus failing the first test of
discontinuity. If we continue to the second step and evaluate its left- and right-sided limits, we see that
9
lim− = −∞
x→3 x − 3
9
lim+ = +∞
x→3 x − 3

meaning the function exhibits an infinite discontinuity.

3 Evaluating limits from a graph — The Organic Chemistry Tutor

(https://youtu.be/7Q2HwTHcxA0)
This video easily shows how to calculate a limit from a graph. According to the video, you will just need
to look at or trace the graph in the direction of the limit being asked, whether it be from the left or from
the right. If both sides give the same limit, the limit exists; otherwise, it doesn’t. The function f (x) merely
checks if a point on the graph is defined.

2
 lim f (x) = 3 lim f (x) = 3 lim f (x) = 3 lim f (x) = 3
x→3− x→3− x→3− x→−2−
lim f (x) = −2 lim f (x) = −2 lim f (x) = −2 lim f (x) = 3
x→3+ x→3+ x→3+ x→−2+
lim f (x) = DN E lim f (x) DN E lim f (x) = DN E lim f (x) = 3
x→3 x→3 x→3 x→−2
f (x) = 1 f (3) = 1 f (3) = 1 f (−2) = 1

lim f (x) = 3 lim f (x) = 1 lim f (x) = 3 π

x→3− x→1− x→3− lim f (x) =
x→−∞ 2
lim f (x) = −2 lim f (x) = 1 lim f (x) = −2 π
x→3+ x→1+ x→3+ lim f (x) =
x→+∞ 2
lim f (x) = DN E lim f (x) = 1 lim f (x) = DN E
x→3 x→1 x→3
f (x) = 1 f (x) = 1 f (3) = 1

lim f (x) = 3 lim f (x) = 3

x→3− x→3−
lim f (x) = −2 lim f (x) = 3
x→3+ x→3+
lim f (x) = DN E lim f (x) = 3
x→3 x→3
f (x) = 1 f (3) = undef ined

4 Evaluating limits by factoring — The Organic Chemistry Tutor

(https://youtu.be/fOrOeZA-vdY)
This video showed another way to evaluate limits, which is by factoring. This is done when the value of
c makes the denominator be 0 (causing the rational function to become undefined) when plugged in. To do
this, we are going to find an expression in the function that is factorable and then proceed by factoring it
out. A simplified expression would then be formed, and we can just use direct substitution to evaluate its
limit.

6x2 + x − 2 (3x + 2)(2x − 1)

2
x − 16 (x + 4)(x − 4) lim1 = lim1
lim = lim x→ 2 2x − 1 x→ 2 2x − 1
x→4 x−4 x→4 x−4
= lim1 3x + 2
= lim x + 4 x→ 2
x→4
7
=8 =
2

x−1 x−1 x2 + x − 20 (x + 5)(x − 4)

lim = lim lim = lim
x→1 x3 − 1 x→1 (x − 1)(x2 + x + 1) x→4 x2 − 5x + 4 x→4 (x − 1)(x − 4)
1 x+5
= lim 2 = lim
x→1 x + x + 1
x→4 x − 1
1 9
= = =3
3 3

x3 + 7x2 + 7x − 15 (x − 1)(x + 3)(x + 5)

lim = lim
x→−5 x+5 x→−5 x+5
= lim (x − 1)(x + 3)
x→−5

= −6(−2) = 12

5 Limits for rational functions — Derek Owens

(https://youtu.be/9neFYGmJp9k)
Any function that can be expressed as a polynomial divided by a polynomial is said to be rational. The
domain of a rational function is the set of all numbers except the zeros of the denominator since polynomials

3
are defined everywhere. Rational functions exhibit specific behaviors that reveal their bounds. In order to
determine if the degree of the numerator function is less than, equal to, or greater than the degree of the
denominator function, we examine the degrees of the quotient functions of rational functions. These traits
will dictate how the boundaries of rational functions behave.

9 9 3x 3(−3) 9 5x + 3 5(−2) + 3 1
lim − = − = −3 lim = =− lim = =
x→3 x 3 x→−3 x2 + 1 (−3)2 + 1 10 x→−2 2x3 + 2 2(−2)3 + 2 2

x5 + x4 − x3 − 1 (x − 1)(x + 1)(x3 − x2 + 1)
lim = lim
x→−1 x+1 x→−1 x+1
x2 + 1 (−1)2 + 1
= lim (x − 1)(x3 − x2 + 1) lim
x→−1 x2 − 1
=
x→−1 (−1)2 − 1
= (−2)(−1) = 2 = DN E

6 Limit of functions at a point — mathdaniwube

(https://youtu.be/q6eNJ6QJhys)
The limit of a function is a value of the function as the input of the function gets closer to or approaches
some number. The video evaluates the limit at a point using the substitution method and expanding through
conjugation method. The substitution may sometimes lead to an indeterminate function because both the
numerator and denominator will equate to 0.

x−2 x−2 x2 − x + 3 (−5)2 + 5 + 3

lim √ = lim lim =
x→2 x − 4x + 4 x→2 x − 2
2 x→−5 1 − 2x 1 − 2(−5)
= lim 1 33
x→2 =
11
=1 =3

x3 − 27 (x − 3)(x2 + 3x + 9)
lim 2
= lim √
x→3 x − 9 x→3 (x − 3)(x + 3) x−4 x−4 x+2
2
x + 3x + 9 lim √ = lim √ ·√
= lim x→4 x − 2 x→4 x − 2 x+2
x→3 x+3 √
(x + 4)( x + 2)
9+9+9 = lim
= x→4 x+4
3+3 √
= lim x + 2
27 9 x→4
= =
6 2 =4

p √ √ p √ √
11 − 4x + 100 − 40 x − x 11 − 2 x − 10 x + 25 − x
lim √ = lim √
x→49 7− x x→49 7− x
√ √
11 − 2( x − 5) − x
= lim √
x→49 7− x
√ √
11 − 2 x + 10 − x
= lim √
x→49 7− x
√
21 − 3 x
= lim √
x→49 7 − x
= lim 3
x→49
=3