NOTE: We cannot apply L'Hopital's Rule since we have not even learned about derivatives yet.
LIMITS
#1. Determine each limit and prove it, using the - definition (LImits Notes, p.1 or Def. 3.1.3 in Lebl).
HINT: See Examples in Limits notes
#1(a).
5𝑥 2 − 5
lim (
𝑥→ 1
#1 (b).
𝑥−1
)
3
lim 𝑥 sin (𝑥)
𝑥→0
Since we know  1  sin
3
 1 for any x  0. Multiply both sides by x for x  0
x
3
x
x
Also, if x  0 then
 x  x sin
3
 x
x
We can handle both cases at once by writing
x  x sin
3
 x
x
Apply the Squeeze theorem:
 x  x sin
3 

lim  x(1)   lim  x sin( )   lim  x(1) 
x 0
x 0
x  x 0

lim  x(1)   0
x 0
lim  x(1)   0
x 0
3 

By the squeeze theorem: lim  x sin( )   0
x 0
x 

3
 0 using the    definition.
x
Begin by letting   0 is given. Find   0 (which depends on  ) so that if 0  x   then
Now we will prove it lim x sin
x 0
f  x   .
Page 1 of 9
Begin with x sin
3
  and solve for x . Then f  x    if
x
x sin
3

x
x sin
3

x

3

 sin x  1


x 3  
x 
Now choose    . Thus, if 0  x   , it follows that f  x    . This completes the proof.
Determine the limit and prove it, using the - definition
#1(c). lim (7𝑥 2 − 13)
𝑥→ 3
Find lim  7 x 2  13 as
x 3
lim  7 x 2  13  7  3  13
2
x 3
 50


Hence lim 7 x2  13  50
x 3
Now we will prove it lim  7 x 2  13  50 using the    definition.
x 3
Begin by letting   0 is given. Find   0 (which depends on  ) so that if 0  x  3   then
f  x   50   .
Begin with f  x   50   and solve for x  3 . Then f  x   50   if
Page 2 of 9
f  x   50  
7 x 2  13  50  
7 x 2  63  
7 x2  9  
7  x  3 x  3  
7 x 3 x 3  
We will now replace the term x  3 with an appropriate constant and keep the term x  3 , since
this is the term we wish to solve for.
To do this, we will arbitrary assume that   1(this is a valid assumption to make since, in
general, once we find a  that works, all smaller values of  will also work ). Then x  3    1
implies that 1  x  3  1 or 2  x  4 so that 5  x  3  7 . It follows that 7 x  3 x  3  
becomes
7 7 x  3  
x 3 

49
  
Now choose   min 1,  . Thus, if 0  x  3   , it follows that f  x   50   . This
 49 
completes the proof.
#2. Find the following limits (if they exist). (You can apply results and theorems in notes and in Lebl; do
not need to work with the definition of limit) Explain/ show work. If a limit does not exist, explain why.
It can be helpful to look at pages 4-6 of my Limits notes. [Note that we have not yet covered differentiation,
and thus cannot apply L'Hopital's Rule.]
𝑥 2 + 𝑥 − 12
#2 (a).
lim
𝑥→ 2 𝑥 2 + 5𝑥 + 4
x 2  x  12
lim 2

x2 x  5 x  4
Plug in the value x  2
x 2  x  12
22  2  12
lim 2
 2
x2 x  5 x  4
2  5 2  4
6

18

1
3
Page 3 of 9
#2 (b).
lim
𝑥 2 + 𝑥 − 12
𝑥→ −4 𝑥 2 + 5𝑥 + 4
x 2  x  12
lim
x 4 x 2  5 x  4
We cannot use the quotient theorem directly because the
limit of the denominator is 0
So, we have to first factor and simplify, and then find the limit:
x 2  x  12 ( x  3)( x  4) ( x  3) 4  3 7 7
lim





x 4 x 2  5 x  4
( x  1)( x  4) ( x  1) 4  1 3 3
x 2  x  12 7

x 4 x 2  5 x  4
3
Hence, lim
#2 (c).
1
lim cos ( )
2x
𝑥→0
Since we know 1  cos
Therefore lim cos
x 0
1
1
 1 for any x  0 , the value of cos oscillates between 1 and 1 .
2x
2x
1
does not exist.
2x
Continuity
#3. Decide if each statement is True or False. If False, provide a counterexample (specifying D and the
functions involved. No proof of true statement requested. Explanation of counterexample not required.)
#3 (a.) FALSE If |f| is continuous on D, then f is continuous on D.
1
Counterexample: f ( x)  
1
The domain is (, )
x0
x0
#3 (b) TRUE If f is continuous on D, then |f| is continuous on D.
#4. For all parts, just state your answers; no explanation required. (You will need to do work to determine
the answers, but you are not required to show it.)
Page 4 of 9
Define 𝑓(𝑥)
= lim
(2 − 𝑥)𝑛
𝑛→∞ 1 + (2 − 𝑥)𝑛
for all real numbers x  3.
(a) To get a feel for how the function behaves, determine each of the following values. That is,
substitute a given x-value, simplify as appropriate, and then find the limit as n  .
f (1) = _____
f (0) = _____
f (1) = _____
f (5/4) = _____
f (2) = _____
f (5/2) = _____
f (4) = _____
f (5) = _____
(b) Determine the numerical value of f(x) for each real number x  3 ----- you should find a
relatively simple multi-part formula for f. Just state your formula.
(c) For what values of x is f continuous? (no explanation required)
Page 5 of 9
#5. Let f: D  R be continuous.
Decide if each statement is True or False. If False, provide a counterexample (specifying D and f; note
that your function f must be continuous.) (No proof of true statement requested. Explanation of
counterexample not required. )
#5 (a) FALSE If D is bounded, then f(D) is bounded.
Let f  x  
1
on D   0,1 . D Is bounded, f  D  is unbounded, because f  x    as x  0
x
#5 (b) FALSE If D is closed, then f(D) is closed.
Let f  x   e x on D   ,   , D is closed, f  D    0,   is open.
#5 (c) TRUE If D is compact, then f(D) is compact. (Recall that compact = closed and bounded.)
#6. Prove that 2x = 7 sin(2x) for some x in the interval (/4, /2). (You can assume the cosine function is
continuous.)
HINT: Let f(x) = 2x  7 sin(2x), and show that f(x) = 0 for some x in (/4, /2). Verify in writing
how the Intermediate Value Theorem can be applied to f on the interval [/4, /2] to produce the
desired result.
Let f  x   2x  7sin 2x
Now we already know polynomial function x is continuous for all x , also trigonometric function
sin3x is continuous for all x . Therefore f  x   2x  7sin 2x is continuous for all x (and
  
obviously in the interval  ,  )
4 2
Here
   2
 
f  
 7 sin 2  
4 4
4



 
 7 sin  
2
2

7
2
 5.43
and

 
 
f    2  7 sin 2  
2
2
2
    0

 3.14
Page 6 of 9




Here point  , 5.43  is below the x axis and point  ,3.14  is above the x axis, and since
4

2

  
function f  x  is continuous in the interval  ,  , it must crosses the x axis in the interval
4 2
  
 , .
4 2
  
Thus, using intermediate value theorem we must have f  x   0 , in the interval  ,  and
4 2
therefore
2 x  7 sin 2 x  0
x
Hence x 
7 sin 2 x
2
7 sin 2 x
  
holds true for some x in the interval  ,  .
2
4 2
Page 7 of 9
Uniform Continuity
#7. Each of the following functions f is continuous on the given set D. Determine whether f is uniformly
continuous on D. Justify your answer by providing a brief explanation (just a sentence or two, no formal proofs).
HINTS: It is helpful to go through the *Theorem and example on page 10 of the LimitsContinuity notes.
#7(a)
𝑓(𝑥) =
8𝑥
D = [1/10, 10]
𝑥
True. Functions f is continuous on the given set D
#7(b)
𝑓(𝑥) =
8𝑥
D = (0, 10]
𝑥
8x
False. The function is not continuous at x  0 , as lim   , hence the function is not bounded
x 0 x
and hence it is not uniformly continuous.

#7 (c)
𝑓(𝑥) =
2𝑥 2 − 18
𝑥−3
D = (0, 3)
True. Functions f is continuous on the given set D
3
𝑥
#7 (d) 𝑓(𝑥) = 𝑥 sin ( )
D = (0, 𝜋)
3
False. The function is not closed and the value of x sin oscillates between 1 and 1, hence it is
x
not uniformly continuous.
Limits at Infinity and Infinite Limits
#8. Let f: (0, )  R. The goal is to use the limit definitions in section 3.5 (Lebl) to prove:
If lim 𝑓(𝑥) =  , then lim
1
𝑥→∞ 𝑓(𝑥)
𝑥→∞
= 0.
Fill in the blanks below. Each __ can be filled in with < or >, as appropriate.
Discussion:
Referring to Def. 3.5.1, suppose  > 0.
1
We want to find real number M such that |𝑓(𝑥) − 0| _<_  whenever x  M.
This means that we want
1
|𝑓(𝑥)|
1
_<_  or equivalently |𝑓(𝑥)| _>_  .
Applying Def. 3.5.6 (Lebl):
Page 8 of 9
Since lim 𝑓(𝑥) = ,
𝑥→∞
1
given 𝑁 =  , there must exist a real number M such that 𝑓(𝑥) _>_ 𝑁 =
Note that for x  M, we must have |𝑓(𝑥)| = 𝑓(𝑥) since 𝑓(𝑥) _>_
1

1

whenever x  M.
1
and  _>_ 0.
Proof:
Let f: (0, )  R. Suppose lim 𝑓(𝑥) = .
𝑥→∞
1

Let  > 0. Set 𝑁 = .
Since lim 𝑓(𝑥) = ,
𝑥→∞
by Def. 3.5.6 (Lebl), there exists real number M such that |𝑓(𝑥)| = 𝑓(𝑥) _>_ 𝑁 =
1

whenever x  M.
Thus, for any x  M,
|
1
𝑓(𝑥)
1
− 0| = |𝑓(𝑥)| =
1
𝑓(𝑥)
_<_
1
1
()
= 𝜀.
Hence Definition 3.5.1(Lebl) holds and lim
1
𝑥→∞ 𝑓(𝑥)
= 0.
Page 9 of 9