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Summer Assignment for BC Calculus
Part 1: A Preview of Calculus and Limits at Infinity
What is Calculus? Use the QR code or follow the link to watch a brief video introduction to
calculus. There is a bit about calculus in multiple dimensions towards the
end though we won’t cover that in this course, it is intersting.
https://www.youtube.com/watch?v=w3GV9pumczQ&vl=en
Now watch a video Introduction to the Concept of a Limit:
https://www.youtube.com/watch?v=U1PxqDjVfBQ
Limits at Infinity
Watch my video for the notes:
Ex 1. Find the infinite limits, limits at infinity, and asymptotes for the function f whose graph is shown:
http://bit.ly/2IbZb3i or
2
Find the limit algebraically. Check your answer graphically on your calculator.
Ex 2. 323
175lim
23
3
++
++
→ xx
xx
x
Ex 3.
2
3x
x 2x 1lim
3x 4x 7→
− +
+ +
Use your calculator for the next one.
Ex 4. a) xx
xx
x 52
16lim
2
3
−
++
→
b)
3
2x
x 6x 1lim
2x 5x→−
+ +
−
Ex. 4 is an example of an infinite limit at infinity. The graph approaches a line called an oblique asymptote.
The equation of the line is the quotient found using polynomial long division.
There is a quick method for finding limits at infinity (i.e. end behavior) of a rational function. As x →∞ or -∞,
the highest degree term of a polynomial function dominates the end behavior of the function. Therefore, the end
behavior of a rational function can be determined as follows:
• degree of the numerator < degree of the denominator →
• degree of the numerator = degree of the denominator →
• degree of the numerator > the degree of the denominator →
3
Some functions can have one HA as x →∞ and a different HA as x → -∞. Finding them algebraically can be
tricky, so we will find them graphically on our calculator.
Ex 5. Find the horizontal asymptotes of 29 1
( )2 5
xf x
x
+=
− .
Ex 6. AB85 2
2n
4nlim
n 10,000n→=
+
A. 0 B. 1
2500 C. 1 D. 4 E. nonexistent
Ex 7. AB93 2
3 2n
3n 5nlim
n 2n 1→
−=
− +
A. -5 B.-2 C. 1 D. 0 E. nonexistent
Ex 8. FRQ 1998 AB2 Let f be the function given by 2xf (x) 2xe= . Find xlim f (x)→−
and xlim f (x)→
graphically. (This was a non-calculator allowed problem on the AP exam.)
Note: x
xlim e 0→−
= and x
xlim e→
= .
4
Practice: Find the Limit
1. x
xlim
x 8→ − 2.
x
2xlim
x 3→ −
3. 2
2h
h 4lim
3h 4→
+
− 4.
3
3x
7p 1lim
2p p→
+
+
5. nlim n→
6. 2
xlim( x 3x 1 x)→
+ + −
7. x
1 xlim
1 x→
−
+ 8. lim cos
→−
9. 2
x
6t 5tlim
(1 t)(2t 3)→−
+
− − 10.
2x
xlim e−
→
11. Horizonatal Asymptote of 2
2
4x x
5x
− 12. Vertical Asymptote of
2x 9
(x 3)(x 4)
−
− −
1
L
a
Summer Assignment for BC Calculus
Part 2: Finding Limits Graphically and Numerically,
One-Sided Limits
http://bit.ly/2K2b5QL
There are three general cases where a limit of a function y = f(x) exists (i.e., equals L) at a value of x = a.
1: The function is continuous at a, as in graph (a) below. It is important to understand that Continuity ensures
the existence of a limit but the converse is not always true:
If the function is continuous at a, then the limit exists at a.
Converse: If the limit exists at a, then the function is continuous. The existence of the limit is a necessary but
not sufficient requirement for continuity.
2 and 3: There is a removable discontinuity at a, as in graphs (b) and (c). The limit still exists at a even though
the function is not continuous at a, because the values of f(x) get closer and closer to L as x approaches a. In
fact, in graph (b), the function is not even defined at a, but the limit still exists.
(a) (b) (c)
Why is the limit L in all three cases? We only care about the y value that the function is approaching as x
approaches a, not what actually happens at f (a).
In this lesson we will use numerical and graphical methods to find lim ( )x a
f x L→
= . In the next lesson we will find
limits analytically using the rules of algebra and theorems of limits.
Ex 1. Evaluate2
lim (3 2)x
x→
− . First enter Y1 = 3x-2 in your calculator.
Graphically: Sketch the graph of y = 3x-2.
Numerically: Use 2nd TBLSET to set TblStart = 1.97, ∆Tbl = .01.
Then use 2nd GRAPH to see the table of values.
2
lim (3 2)x
x→
− = _____.
L
a
L
a
2
Ex 2. Evaluate21
1lim
1x
x
x→
−
−. First enter Y1 =
2
1
1
x
x
−
− in your calculator.
Graphically: Sketch the graph of f(x) = 2
1
1
x
x
−
−.
Note that there is a removable discontinuity at x =1.
(Use ZOOM 4 to see it on your calculator.)
Numerically: Use the TRACE key to complete the tables to 6 places.
Let’s look at f (x) as x approaches 1 from the left, getting closer and closer, but never equaling 1.
x 0.5 0.9 0.99 0.999 0.9999
f (x)
Let’s look at f (x) as x approaches 1 from the right, getting closer and closer, but never equaling 1.
x 1.5 1.1 1.01 1.001 1.0001
f (x)
Does f (1) exist? _____ Explain: __________________________________________________.
To what value does f (x) approach from the right and the left? _________ 21
1lim
1x
x
x→
−
− = _____.
Now let’s change f(x) slightly by giving it a value of 2 when x = 1 and calling it g(x). Find the same limit.
2
11
( ) 1
2 1
xif x
g x x
if x
−
= − =
Graphically: Sketch the graph of g(x).
Note that there is different kind of removable discontinuity at x =1.
Numerically: The above table also applies here.
Does g (1) exist? _____ Explain: __________________________________________________.
To what value does f (x) approach from the right and the left? _________ 21
1lim
1x
x
x→
−
− = _____.
3
1. Answer each using the graph below
a. ( ) =
−→xf
x 5lim
b. ( ) =
→xf
x 2lim
c. ( ) =
→xf
x 0lim
d. ( ) =
→xf
x 4lim
2. Using a graph or table find 0
sinlimx
x
x→. Hint: graph y =
sin x
x
using ZOOM 7. Then find 0
sinlimx
x
x→
either graphically using ZOOM 7 and TRACE, or numerically using 2nd TBLSET to set TblStart = 0 and
∆Tbl = .01.
_____________________________________________________________________________________
Practice:
1. Find 3
2lim
3x x→ − using graphical or numerical methods.
2. (1997AB15)
The graph of the function f is shown in the figure above. Which of the following statements about f is true?
(A)
limx→af (x)= lim
x→bf (x) (B)
limx→ af (x)= 2 (C)
limx→ bf (x)= 2
(D)
limx→bf (x)= 1 (E)
limx→af (x) does not exist
0
sinlimx
x
x→ = ____________.
4
3. Explain and make a sketch of each of the three scenarios in which a limit does not exist at a. Use the internet
(google images, etc.) for help.
Case 1: Discontinuity (Jump) Case 2: Unbounded Behavior Case 3: Oscillating Behavior
4: Show the x 0
| x |lim
x→does not exist
5. Refer to the diagram.
a. f (-3) b. x 3lim f (x)
−→− c.
x 3lim f (x)
+→−
d. x 3lim f (x)→−
e. f (-1) f. x 1lim f (x)
−→−
g. x 1lim f (x)
+→− h.
x 1lim f (x)→−
i. f (2)
j. x 2lim f (x)
−→ k.
x 2lim f (x)
+→ l.
x 2limf (x)
→
5
4. The graph of the function g is shown. Use it to state the values (if they exist) of the following:
a) 2
lim ( )x
g x−→
b) 2
lim ( )x
g x+→
c) 2
lim ( )x
g x→
d) 5
lim ( )x
g x−→
e) 5
lim ( )x
g x+→
f) 5
lim ( )x
g x→
5. Use the graph to answer a- i:
a. 2
lim ( )x
f x+→−
= b. 2
lim ( )x
f x−→−
= c. 2
lim ( )x
f x→−
=
d. 2
lim ( )x
f x+→
= e. 2
lim ( )x
f x−→
= f. 2
lim ( )x
f x→
=
g 0
lim ( )x
f x+→
= h. 0
lim ( )x
f x−→
= i. 0
lim ( )x
f x→
=
6.
7. Find where
8. Decide if the following are sometimes, always, or never true.
a. If 2
lim ( )x
f x→
does not exist, then 2
lim ( )x
f x+→
does not exist.
b. If 2
lim ( )x
f x+→
does not exist, then 2
lim ( )x
f x→
does not exist.
=−
−−→ 3
3lim
3 x
x
x
( )
+
=
+
=
23
28
213
2 xx
x
xx
xf( )2
limx
f x→
Summer Assignment BC Calculus
Part 3 - Finding Limits Analytically
Sometimes we will want to evaluate a limit using an analytical or algebraic method. As we saw before,
sometimes the numeric methods and graphing methods don’t work out well. There are essentially three ways to
do this: substitution, factoring and substitution, using the conjugate and substitution.
We also need to know the properties of limits below to help us evaluate limits:
Let b and c be real numbers, let n be a positive integer, and let f and g be functions with the limits lim𝑥→𝑐
𝑓(𝑥) = 𝐿
and lim𝑥→𝑐
𝑔(𝑥) = 𝐾 .
1. Scalar Multiple: lim𝑥→𝑐
[𝑏𝑓(𝑥)] = 𝑏𝐿
2. Sum or Difference: lim𝑥→𝑐
[𝑓(𝑥) ± 𝑔(𝑥)] = 𝐿 ± 𝐾
3. Product: lim𝑥→𝑐
[𝑓(𝑥)𝑔(𝑥)] = 𝐿𝐾
4. Quotient: lim𝑥→𝑐
𝑓(𝑥)
𝑔(𝑥)=
𝐿
𝐾, 𝐾 ≠ 0
5. Power: lim𝑥→𝑐
[𝑓(𝑥)]𝑛 = 𝐿𝑛
Limits of Polynomial and Rational Functions
If p is a polynomial function and c is a real number, then
lim𝑥→𝑐
𝑝(𝑥) = 𝑝(𝑐)
If r is a rational function given by 𝑟(𝑥) =𝑝(𝑥)
𝑞(𝑥) and c is a real number such that 𝑞(𝑐) ≠ 0, then
lim𝑥→𝑐
𝑟(𝑥) = 𝑟(𝑐) =𝑝(𝑐)
𝑞(𝑐)
The Limit of a Function Involving a Radical
Let n be a positive integer. The limit below is valid for all c when n is odd and is valid for 𝑐 > 0 when n is
even.
lim𝑥→𝑐
√𝑥𝑛
= √𝑐𝑛
The Limit of a Composite Function
If f and g are functions such that
lim𝑥→𝑐
𝑔(𝑥) = 𝐿 and lim𝑥→𝐿
𝑓(𝑥) = 𝑓(𝐿)
then
lim𝑥→𝑐
𝑓(𝑔(𝑥)) = 𝑓(lim𝑥→𝑐
𝑔(𝑥)) = 𝑓(𝐿)
Video explanation of the Limit Laws
http://bit.ly/2Xvx7hV
or
Limits of Transcendental Functions
Let c be real number in the domain of the given transcendental function. 1. lim
𝑥→𝑐sin 𝑥 = sin 𝑐
2. lim𝑥→𝑐
cos 𝑥 = cos 𝑐
3. lim𝑥→𝑐
tan 𝑥 = tan 𝑐
4. lim𝑥→𝑐
cot 𝑥 = cot 𝑐
5. lim𝑥→𝑐
sec 𝑥 = sec 𝑐
6. lim𝑥→𝑐
csc 𝑥 = csc 𝑐
7. lim𝑥→𝑐
𝑎𝑥 = 𝑎𝑐 , 𝑎 > 0
8. lim𝑥→𝑐
ln 𝑥 = ln 𝑐
Practice Using Limit Laws:
Ex 1: Find →
−
−31
2 1lim
2x
x
x
Ex. 2: Find given
You try: Use the applicable Limit Laws to find the limits if →
=lim ( ) 2x c
f x and →
= −lim ( ) 3x c
g x .
a) b)
c) d)
e) f)
( )( )xgfx
+→2
lim
lim 2 ( ) 4 ( )x c
f x g x→
− 3
lim ( ) 1x c
f x→
+
2 ( ) 3 ( )lim
( ) ( )x c
f x g x
g x f x→
+
− lim ( ) 2 ( )
x cf x g x
→
( )lim ( ) ( ) 3x c
f x g x→
+
2( )lim
3 ( )x c
f x
g x→ − −
Substitution:
If you have a known continuous function (for example, all polynomials are continuous) or a function that you
know is continuous at a, you can simply substitute the value that x is approaching into the function and
evaluate. Check out these examples:
Ex 1: lim𝑥→4
5𝑥 + 3 Ex 2: lim𝑥→−2
4𝑥+9
𝑥−2
Factoring and Substitution:
Ex 3: lim𝑥→−2
(𝑥2+7𝑥+10)
𝑥+2 Ex. 4: lim
𝑥→−2
(𝑥−2)2−16
𝑥+2
Rationalization and Substitution:
Ex. 5: lim𝑥→0
√𝑥+25−5
𝑥 Ex. 6: lim
𝑥→0
√3+𝑥−√3
𝑥
Fractions and Substitution:
Ex. 7: lim𝑥→0
1
3+𝑥−
1
3
𝑥
For more examples of these techniques, check out the following clip: www.calculus-help.com/tutorials lesson 3.
Squeeze Theorem:
If ℎ(𝑥) ≤ 𝑓(𝑥) ≤ 𝑔(𝑥) for all x in an open interval containing c, except possibly at c itself, and if
lim𝑥→𝑐
ℎ(𝑥) = 𝐿 = lim𝑥→𝑐
𝑔(𝑥)
then lim𝑥→𝑐
𝑓(𝑥) exists and is equal to L.
Example of the Squeeze Theorem in action… http://bit.ly/31j7nr0 or
Three Special Limits:
1. lim𝑥→0
sin 𝑥
𝑥= 1 2. lim
𝑥→0
1−cos 𝑥
𝑥= 0 3. lim
𝑥→0(1 + 𝑥)
1𝑥⁄ = 𝑒
Proof of the first special limit using the Squeeze Theorem:
http://bit.ly/2KF910e
Practice with Special Limits
Ex 1: lim𝑥→0
sin 𝑥
4𝑥 Ex 2: lim
𝑥→0
(sin 𝑥)2
𝑥
Ex 3: lim𝑥→𝜋
2⁄
cos 𝑥
cot 𝑥 Ex 4: lim
𝑥→0
1−𝑒−𝑥
𝑒𝑥−1
1
Continuity at a point: A function f is continuous at a number c if
lim ( ) ( )x c
f x f c→
= .
This definition implicitly requires three conditions for f(x) to be
continuous at c:
Summer Assignment for BC Calculus
Part 4: Continuity
Ex 1: Study the graph below and determine where the function is discontinuous. Identify the removable
discontinuities and define the value of the function in order to remove the discontinuity.
1
1
Ex 2: Where are each of the following functions discontinuous? State the type of discontinuity.
a. 2 2
( )2
x xf x
x
− −=
− b. 2
1, if 0
( )
1, if 0
xf x x
x
= =
c.
2 2, if 2
( ) 2
1, if 2
x xx
f x x
x
− −
= − =
d. ( )f x x=
Solution:
http://bit.ly/2EYBZoj
2
To summarize what we saw in the examples, there are 3 types of discontinuity at a point where x = c:
1. Removable Discontinuity (RD): A limit exists at x = c but there is a hole at x = c. The
discontinuity can be removed by redefining the function to equal the limit’s value at x = c.
2. Jump Discontinuity (JD): A limit DNE; the left and right hand limits exist but are not equal.
3. Infinite Discontinuity (ID): A limit DNE; the left and right hand limits each equal or − .
Removable Discontinuities
The function p(x) above has a removable discontinuity at x = 3 because the function can be made
continuous by appropriately defining the function at the discontinuity, i.e., (3) 6p = .
2 93
( ) 3
6 3
xif x
p x x
if x
−
= −
=
REMOVABLE discontinuities & holes both have limits and are basically the same thing.
Jumps and infinite discontinuities do not have a limit and are NON-REMOVABLE.
Practice: Continuity:
1) Using the three part definition of continuity, determine if the function is continuous at the given
point.
a. ( )f x x= at x = 0.
b. 2 16
( ) at 44
xf x x
x
−= =
−
c. ( ) at 1x
f x xx
= =
d. 2( ) 1 at 0f x x x= + =
3
e.
0
( ) at 0
0
xx
f x xx
x x
= = =
f.
2 42
( ) at 22
4 2
xx
f x xx
x
−
= =− =
g.
11
1( ) 1 1 at 1
2
1
xx
xf x x x
x x
−
−
= − = −
h.
11
1( ) 1 1 at 1
2
1
xx
xf x x x
x x
−
−
= − =
2) a. Is ( )f x x= continuous on (1, 2)?
b. Is ( ) 1f x x= − continuous on [1, 2]?
4
3) a. At which points is the graph discontinuous?
b. On what intervals is the graph continuous?
The following theorems allow you to determine if combinations of continuous functions are continuous.
(1) If f and g are both continuous at a point a, and k is a constant, then so are
( ) and n nf
k f, f + g, f - g, fg, g = 0 , f , f g
.
(2) If f and g are both continuous on an interval [a, b], and k is a constant, then so are
andf
k f, f + g, f - g, fg,g
(except where g = 0)
(3) The following functions are continuous at every number in their domains (where they are defined):
(a) Polynomials
(b) Rational functions
(c) Root functions
(d) Trigonometric functions
(e) Exponential functions
(f) Inverses of any continuous functions
Some useful results of these theorems are:
(a) A polynomial function is continuous for all real numbers.
(b) A rational function is continuous for all real numbers except at a hole or vertical asymptote.
(c) The absolute value function is continuous for all real numbers.
(d) The nth root function ( ) nf t x= is continuous at every real number if n is odd, and at every real
number in [0, ) is if n is even.
(e) If f(x) and g(x) are continuous for all real numbers x, then f(g(x)) is continuous on all real numbers. If
only f(x) is continuous all real numbers, then f(g(x)) is continuous on the domain of g(x).
1
Summer Assignment BC Calculus
Part 5: Intermediate Value Theorem and Infinite Limits
Intermediate Value Theorem (IVT)
Check out this video for an introduction:
https://www.youtube.com/watch?v=6AFT1wnId9U
INTERMEDIATE VALUE THEOREM (IVT)
Suppose that f is continuous on the closed interval ,a b and let k be any number between
f(a) and f(b) (i.e., f(a) < k < f(b) ). Then there exists a number c in (a, b) such that f(c) = k.
In geometric terms, the IVT says that if the graph of a function between x = a and x = b has no hole, break, or
vertical asymptote, then the function hits every y-value between y = f(a) and y = f(b).
A real world example is if you are on a road traveling at 40 mph and a minute later you are traveling at 50
mph, at some point within that minute, you must have been traveling at 41 mph, 42 mph, and every
possible value between 40 mph and 50 mph.
Watch my video to see some of the following examples worked out…
http://bit.ly/2XyjoGP or
Determine if the Intermediate Value Theorem holds for the given function, interval, and value of k. If the
theorem holds, find a number c in (a,b) such that f(c) = k. If the theorem does not hold, give the reason.
Ex 1: 2( ) 5 6; [ , ] [ 1,2]; 4f x x x a b k= + − = − = Ex 2: 4 1
( ) ; [ , ] [ 3,1];2 2
f x a b kx
= = − =+
2
Determine if the Intermediate Value Theorem holds for the given function and interval. If the theorem
holds, find a number c in (a,b) such that f(c) = k. If the theorem does not hold, give the reason.
Ex 3: 4 1
( ) ; [ , ] [ 1,1];2 2
f x a b kx
= = − =+
Ex 4: 2( ) 25 ; [ , ] [ 4.5,3]; 3f x x a b k= − = − =
Ex 5: Show that f(x) = x3 + 2x - 1 has a zero on the interval (0, 1).
Ex 6: Determine if the Intermediate Value Theorem applies in each problem. Explain your answer
a. Currently your height is 5 feet 8 inches. At some time in your life you were 3 feet tall.
b. You travelled 60 miles east from home to work in 1 hour. Your friend says that he can use the
IVT to prove that he saw you 70 miles east from home before you arrived at work.
3
Mixed Practice
1. (1985AB29) Which of the following functions are continuous for all real numbers x?
I.
y = x
2
3 II.
y = ex III.
y = tan x
(A) None (B) I only (C) II only (D) I and II (E) I and III
2. (1993AB5) If the function f is continuous for all real numbers and if
f (x) =x
2− 4
x + 2 when x −2 ,
then
f (−2) =
(A) –4 (B) –2 (C) –1 (D) 0 (E) 2
3. (1998AB26)
x 0 1 2
f (x) 1 k 2
The function f is continuous on the closed interval
0,2 and has values that are given in the table above. The
equation
f (x) =1
2 must have at least two solutions in the interval
0,2 if
k =
(A)
0 (B)
1
2 (C) 1 (D) 2 (E) 3
4. (1988BC5) Let f be the function defined by
f (x) =
sin x if x 0
x2 if 0 x 1
2 − x if 1 x 2
x − 3 if x 2
For what values of x is f NOT continuous?
(A) 0 only (B) 1 only (C) 2 only (D) 0 and 2 only (E) 0, 1, and 2
5. (2003AB4) Let f be the function defined by
1 for 0 3
( )5 for 3 5
x xf x
x x
+ =
−
Is f continuous at 3x = ? Explain why or why not
4
6. Find a number k that makes f(x) continuous at x = 5.
2 2 5( )
4 7 5
kx if xf x
x if x
+ =
+
Strategy:
Using the three part definition of continuity, discuss the continuity at a point c.
7a. 1
( ) 1f x x at cx
= − = 7b. 2 4
( ) 22
xf x at c
x
−= =
−
8. 2
2 2( ) 2
2
x if xf x at c
x x
− = =
9.
3 11
( ) 1
3 1
xfor x
f x x
for x
−
= −
=
In these three exercises, the given function has a removable discontinuity at a certain point. How should
the function be defined in order to make it continuous everywhere?
10. 24 9
( )2 3
xf x
x
−=
− 11.
4( )
2
xf x
x
−=
−
5
12. Consider the pictured function:
a. Where on the interval 1 7x− does the limit of the function fail to exist?
b. Where on the interval 1 7x− does the function fail to be continuous?
6
QR Codes/Links to filled in Summer assignments.
Part 1: Part 2:
http://bit.ly/2R6VujB http://bit.ly/2ZjRNJU
Part 3: Part 4:
http://bit.ly/2MIkkHK http://bit.ly/2WDJoo0
Part 5:
http://bit.ly/2IzV9lt