Math 1210 § 4.
Treibergs
First Midterm Exam
Name: Practice Problems
September 4, 2015
Practice Questions. The upcoming exam questions may be similar to these, but not identical.
Understand the concepts so that you handle minor variations.
1. Determine whether the indicated limits exist. If they do, find them. Be sure to show all
work.
x2 − x − 6
x→3 x2 − 2x − 3
x2 + x
(b) lim 2
x→4 x + 2x + 3
(c) Recall that [[x]] denotes the greatest integer part of x.
lim [[x2 + 1]]
(a) lim
x→5
2. Sketch the graph of


x + 1, if x < 0;


if 0 ≤ x < 1;
f (x) = x,

1

 ,
if 1 ≤ x.
x
Find each of the following limits or state that it does not exist. Explain.
(a)
lim f (x).
x→−∞
(b) lim f (x).
x→0
(c) lim f (x).
x→0−
(d) lim f (x).
x→1
(e) lim f (x).
x→∞
3. Determine whether the indicated limits exist. If they do, find them. Be sure to show all
work.
√
√
x+3− 3
(a) lim
x→0
x
x−4
(b) lim
x→4 |x − 4|
1
(c) lim x2
x→0
x2
4. Recall the Main Limit Theorem
Theorem A. Main Limit Theorem. Let n be a positive integer, k be a constant and f
and g functions that have a limit at c. Then
1. lim k = k;
x→c
2. lim x = c;
x→c
3. lim kf (x) = k lim f (x);
x→c
x→c
4. lim f (x) + g(x) = lim f (x) + lim g(x);
x→c
x→c
x→c
1
5. lim f (x) − g(x) = lim f (x) − lim g(x);
x→c
x→c
x→c
6. lim f (x) · g(x) = lim f (x) · lim g(x) ;
x→c
x→c
x→c
lim f (x)
f (x)
= x→c
, provided lim g(x) 6= 0;
x→c
x→c g(x)
lim g(x)
x→c
n
n
8. lim (f (x)) = lim f (x) ;
x→c
x→c
q
p
n
n
9. lim f (x) =
lim f (x), provided lim f (x) > 0 when n is even.
7. lim
x→c
x→c
x→c
Use Theorem A to find the limit. Justify each step by appealing to a numbered statement.
p
lim −2w3 + 7w2 .
x→2
5. Determine whether the indicated limits exist. If they do, find them. Be sure to show all
work.
tan 2x
tan 3x
θ cot θ
(b) lim
θ→0 sec θ
1 − cos t
(c) lim
t→0 sin2 t
(a) lim
x→0
6. Determine whether the indicated limit exists. If it does, find it. Be sure to show all work.
lim (−1)
hh ii
1
x
x→0
sin(x)
7. Determine whether the indicated limits exist. if they do, find them. Be sure to show all
work.
(2x + 3)4
x→∞ x4 + 1
x+2
(b) lim 3
x→1 x − 1
x2 + 3x
(c) lim
x→2 (x − 2)4

 sin x
, if x 6= 0;
8. Let f (x) =
x
1,
if x = 0.
(a) lim
Determine whether f (x) is continuous on the interval [0, 2].
9. A particle moves along a coordinate line and s, its
√ directed distance in centimeters from
the origin after t seconds is given by s = f (t) = 7t + 1. Using just the limit definition,
find the instantaneous velocity of the particle after 5 seconds.
10. Using just the limit definition, for each x, determine whether F (x) is differentiable at x,
1
and if it is, find F 0 (x), where F (x) =
.
x+2
2
Solutions.
1. Determine whether the indicated limits exist. If they do, find them. Be sure to show all
work.
(a)
(x − 3)(x + 2)
(x + 2)
3+2
x2 − x − 6
5
= lim
= lim
=
=
.
2
x→3 (x − 3)(x + 1)
x→3 (x + 1)
x→3 x − 2x − 3
3+1
4
lim
(b)
20
42 + 4
x2 + x
=
=
2
2
x→4 x + 2x + 3
4 +2·4+3
27
lim
(c) lim [[x2 + 1]] Note that [[y]] jumps whenever y is an integer. Also we see that lim x2 + 1 =
x→5
x→5
52 + 1 = 26. Since x2 + 1 is an increaseing function, we see that the function
f (x) = [[x2 + 1]] equals 26 for x’s close to but greater than 5, and f (x) = 24 for
x’s close to but less than 5. Because f (x) jumps at x = 5, the limit does not exist .
2. Sketch the graph of
−0.5
0.0
y
0.5
1.0
1.5


x + 1, if x < 0;


if 0 ≤ x < 1;
f (x) = x,

1

 ,
if 1 ≤ x.
x
−1.0
−0.5
0.0
0.5
x
3
1.0
1.5
2.0
Find each of the following limits or state that it does not exist. Explain.
(a)
lim f (x) = −∞ . f (x) decreases to −∞ without bound as x → −∞.
x→−∞
(b) lim f (x)
x→0
Does not exist. . The function jumps at x = 0. The function approaches
different values for x > 0 and x < 0 so there is no consistent limiting value.
(c) lim f (x) = 1 . The function approaches one as x → 0 from the left.
x→0−
(d) lim f (x) = 1 . The left and right limits have the same limiting value, so the two-sided
x→1
limit exists at x = 1.
(e) lim f (x) = 0. The quantity 1/x decreases to zero as x → ∞.
x→∞
3. Determine whether the indicated limits exist. if they do, find them. Be sure to show all
work.
(a)
√
lim
x→0
x−3−
x
√
3
√
= lim
x→0
√ √
√ x+3− 3
x+3+ 3
√ √
x x+3+ 3
(x + 3) − 3
√ √
x x+3+ 3
x
√ = lim
√
x→0 x
x+3+ 3
= lim
x→0
= lim √
x→0
1
1
√ = √
x+3+ 3
2 3
x−4
x−4
equals +1 for x > 4 and
Does not exist. The function f (x) =
|x − 4|
|x − 4|
−1 for x < 4, thus has a jump at x = 4. The two sided limit does not exist because
f (x) has differing limits upon left and right approaches x → 4.
1
2
(c) lim x
= 1.
x→0
x2
1
This can be seen from the Squeeze Theorem. Note that as x → 0 the fraction 2 tends
x
ti infinity without bound. The greatest integer part drops the fractional part of this
value, thus is at most one smaller, and so satisfies the inequality
1
1
1
−1≤
≤ 2
x2
x2
x
1
Multiplying by x2 , we see that the function f (x) = x2
satisfies
x2
(b) lim
x→4
1 − x2 ≤ f (x) ≤ 1
so that in the limit, because the first and last expressions tend to one, the middle
quantity does also by the Squeeze Theorem.
4
4. Recall the Main Limit Theorem
Theorem A. Main Limit Theorem. Let n be a positive integer, k be a constant and f
and g functions that have a limit at c. Then
1. lim k = k;
x→c
2. lim x = c;
x→c
3. lim kf (x) = k lim f (x);
x→c
x→c
4. lim f (x) + g(x) = lim f (x) + lim g(x);
x→c
x→c
x→c
5. lim f (x) − g(x) = lim f (x) − lim g(x);
x→c
x→c
x→c
6. lim f (x) · g(x) = lim f (x) · lim g(x) ;
x→c
x→c
x→c
lim f (x)
f (x)
= x→c
, provided lim g(x) 6= 0;
x→c
x→c g(x)
lim g(x)
x→c
n
n
8. lim (f (x)) = lim f (x) ;
x→c
x→c
q
p
n
9. lim f (x) = n lim f (x), provided lim f (x) > 0 when n is even.
7. lim
x→c
x→c
x→c
Use theorem A. to find the limit. Justify each step by appealing to a numbered statement.
q
p
lim −2w3 + 7w2 = lim −2w3 + 7w2
by (9). We’re taking the n = 2nd root, which
x→2
x→2
is even, but inner limit is positive, as we shall see.
=
q
=
q
by (4), the limit of a sum is the sum of a limit.
−2 lim w3 + 7 lim w2
by (3), a constant multiple may be pulled out.
x→2
x→2
r
=
=
lim −2w3 + lim 7w2
x→2
x→2
2
3
−2 lim w + 7 lim w
x→2
p
−2 · 23 + 7 · 22 =
x→2
√
√
12 = 2 3 .
by (8), limit of power is the power of limit.
by (2), limit of x itself is c.
5. Determine whether the indicated limits exist. If they do, find them. Be sure to show all
work.
(a)
sin 2x
2
tan 2x
cos 2x
2x
lim
= lim x→0 tan 3x
x→0
3
sin 3x
cos 3x
3x
2
sin 2x
lim
lim
x→0 cos 2x
x→0 2x
=
3
sin 3x
lim
lim
x→0 cos 3x
x→0 3x
2
sin y
lim
lim
y→0 cos y
y→0 y
=
where we let y = 2x and z = 3x.
3
sin z
lim
lim
z→0 cos z
z→0 z
5
So
2
(1)
2
tan 2x
1
lim
= =
.
x→0 tan 3x
3
3
(1)
1
(b)
θ cot θ
θ cos2 θ
= lim
θ→0 sec θ
θ→0 sin θ
cos2 θ
= lim θ→0
sin θ
θ
2
lim cos θ
θ→0
12
=
= = 1.
sin θ
1
lim
θ→0 θ
lim
(c)
lim
t→0
1 − cos t
1 − cos t
= lim
t→0 1 − cos2 t
sin2 t
1 − cos t
= lim
t→0 (1 − cos t) (1 + cos t)
1
= lim
t→0 1 + cos t
lim 1
1
1
t→0
=
=
=
.
lim 1 + lim cos t
1+1
2
t→0
t→0
6. Determine whether the indicated limit exists. If it does, find it. Be sure to show all work.
hh ii
1
x
lim (−1)
x→0
sin(x) = 0 .
1
tends to
This can be seen from the Squeeze Theorem. Note that as x → 0 the fraction
x
plus or minus infinity without bound. The greatest integer part drops the fractional part of
this value, and so also tends to plus or minus infinity. Raising −1 to this power oscillates
between ±1. Thus we have a bound
hh ii 1 (−1) x ≤ 1
Multiplying by | sin x|, we see that
hh ii 1 sin(x) (−1) x ≤ | sin x|
which is the same as saying
−| sin x| <
hh ii
1
sin x (−1) x
6
≤ | sin x|.
Since the absolute value function and the sine function are both continuous, the composite
| sin x| is continuous also. Its limit as x → 0 is | sin(0)| = 0. Because the first and last
expressions tend to zero, the middle quantity does also by the Squeeze Theorem.
7. Determine whether the indicated limits exist. if they do, find them. Be sure to show all
work.
(a)
4
3
2+
(2x + 3)4
x
=
lim
lim
1
x→∞
x→∞ x4 + 1
1+ 4
x
4
3
lim 2 +
x→∞
x
=
1
lim 1 + 4
x→∞
x
4
3
lim 2 + lim
x→∞
x→∞ x
=
1
lim 1 + lim 4
x→∞
x→∞ x
4
1
lim 2 + 3 lim
x→∞
x→∞ x
(2 + 3 · 0)4
=
=
= 16 .
1
1+0
lim 1 + lim 4
x→∞
x→∞ x
(b)
lim
x→1
x+2
x+2
= lim
3
x→1
x −1
(x − 1)(x2 + x + 1)
Does not exist .
When x is close to one the numerator x + 2 is close to 3 and x2 + x + 1 is also close
1
to 3. Thus the ratio is approximately x−1
which is close to +∞ or −∞ depending on
whether x > 1 or x < 1. Since there is a jump, the two sided limit does not exist.
x2 + 3x
= ∞
x→2 (x − 2)4
Note that (x − 2)4 stays positive and tends to zero as x → 2. However, for x near 2,
the numerator x2 + 3x is close to 10, therefore, the ratio becomes unboundedly large
for both left and right approaches of x to 2.

 sin x
, if x 6= 0;
8. Let f (x) =
x
1,
if x = 0.
(c) lim
Determine whether f (x) is continuous on the interval [0, 2].
f is continuous on [0, 2] . We need to check that for each c such that 0 ≤ c ≤ 2 we have
lim f (x) = f (c).
x→c
For c = 0 the limit
lim f (x) = lim
x→0+
x→0+
7
sin x
= 1 = f (0).
x
For 0 < c we know that both functions x and sin x are continuous at c and that x is nonzero
at c. Therefore the limit for 0 < c < 2 is
lim sin x
sin x
sin c
= x→c
=
= f (c).
x→c x
lim x
c
lim f (x) = lim
x→c
x→c
Similarly, at the right endpoint c = 2,
sin x
=
x→2− x
lim f (x) = lim
x→2−
lim sin x
x→2−
lim x
x→2−
=
sin 2
= f (2).
2
9. A particle moves along a coordinate line and s, its
√ directed distance in centimeters from
the origin after t seconds is given by s = f (t) = 7t + 1. Using just the limit definition,
find the instantaneous velocity of the particle after 5 seconds.
Inserting into the formula for the instantaneous velocity at t we compute
f (t + h) − f (t)
h
p
√
7(t + h) + 1 − 7t + 1
= lim
h→0
h
p
p
√
√
7(t + h) + 1 − 7t + 1
7(t + h) + 1 + 7t + 1
p
= lim
√
h→0
h
7(t + h) + 1 + 7t + 1
V = lim
h→0
= lim
h→0
[7(t + h) + 1] − [7t + 1]
p
√
h
7(t + h) + 1 + 7t + 1
7h
p
√
h→0
h
7(t + h) + 1 + 7t + 1
= lim
= lim p
h→0
7
7(t + h) + 1 +
√
7t + 1
=
7
√
centimeters per second.
2 7t + 1
10. Using just the limit definition, for each x, determine whether F (x) is differentiable at x,
1
. The function is differentiable if the limit of
and if it is, find F 0 (x), where F (x) =
x+2
the difference quotient exists. Indeed,
F (x + h) − F (t)
h
1
1
−
(x + h) + 2 x + 2
= lim
h→0
h
[x + 2] − [(x + h) + 2
= lim
h→0 h[(x + h) + 2] [x + 2]
−h
= lim
h→0 h[(x + h) + 2] [x + 2]
F 0 (x) = lim
h→0
−1
1
= −
.
h→0 [(x + h) + 2] [x + 2]
(x + 2)2
= lim
The derivative is defined when division by zero does not occur, namely when x 6= −2.
8