PRACTICE PROBLEMS: SET 1
MATH 100: PROF. DRAGOS GHIOCA
1. Problems
Problem 1. Compute the following limits:
(a)
ex−4 − x + 1
lim 2
x→4 x − ln(x − 3)
(b)
(1 + h)2 − 1
lim
h→0
h
(c)
lim |x − 3|
x→3
(d)
lim
x→0
x
|x|
(e)
lim
x→−1+
5x
x+1
(f)
1
lim e x
x→0−
(g)
√
x2 + 9 − 5
lim
x→4
x−4
Problem 2.
(a) Let
{
f (x) =
Compute limx→0 f (x).
(b) Let
g(x) =
{
x, if
x2 , if
5 − x, if
ln(x), if
x<0
x≥0
x<1
x≥1
Compute limx→1 g(x).
Problem 3. The displacement (in meters) of a particle moving along a straight
line is given by the equation of motion s = 3t2 + t, where t is measured in seconds.
(a) Find the average velocity during the time period [1, 5].
(b) Find the instantaneous velocity of the particle when t = 2.
1
2
MATH 100: PROF. DRAGOS GHIOCA
2. Solutions.
Problem 1.
(a) We first try to plug in x = 4 into the given fraction and see if we get
something meaningful. In this case we are lucky and the limit is
simply computed as this; in the other limits from this problem
we won’t be so lucky again. So,
ex−4 − x + 1
e0 − 3
−2
1
=
=
=− .
x→4 x2 − ln(x − 3)
16 − ln(1)
16
8
lim
(b) Here if we plug in h = 0 in the given limit we get “ 00 ” which always requires
more work from us. What we have to do in this case is clear out the
algebra on the top and look for a cancellation of h from both numerator
and denominator. Only after we cancel out h we will be able to plug
in h = 0 in the newly formed expression. So,
(1+h)2 −1
h
2
−1
limh→0 1+2h+h
h
2
= limh→0 h +2h
h
limh→0
=
= limh→0 h + 2
= 2.
(c) In this case, theoretically we could plug in x = 3 in the given limit, however
this would be wrong. The reason for this is that the absolute value function
is a piecewise defined function and thus it always requires us to use lateral
limits when we deal with its branching point.
We recall that
{
−x, if x < 0
|x| =
x, if x ≥ 0
So,
{
|x − 3| =
−(x − 3), if
x − 3,
if
{
and thus
|x − 3| =
3 − x, if
x − 3, if
x−3<0
x−3≥0
x<3
x≥3
Thus
lim |x − 3| = lim− 3 − x = 0
x→3−
x→3
and
lim |x − 3| = lim+ x − 3 = 0.
x→3+
x→3
In conclusion, since the lateral limits are both equal to 0, we conclude that
also the global limit exists and it equals 0, i.e.,
lim |x − 3| = 0.
x→3
(d) As in the previous exercise, we need to consider lateral limits. So,
x
x
= lim−
= lim− −1 = −1
lim
x→0 −x
x→0
x→0− |x|
PRACTICE PROBLEMS: SET 1
3
and
lim+
x→0
x
x
= lim+ = lim+ 1 = 1.
|x| x→0 x x→0
In this case, the lateral limits do not agree, which yields that
lim
x→0
x
does not exist.
|x|
(e) In this case when we try to plug in x = −1 in the given function we get
something of the form “ −5
0 ”. This expression by itself does not make
sense but it suggests the correct answer: the limit does not exist, and actually the limit diverges to either +∞, or −∞. Now, in order to determine
which is the correct divergence point, we note that as x approaches −1, the
numerator gets very close to −5; so, in particular, it is negative. As x approaches −1 from the right (as the limit asks from us), the denominator
approaches 0 through positive values since
x + 1 > 0 as x > −1.
In conclusion, the fraction will always stay negative and thus
lim +
x→−1
5x
diverges to − ∞.
x+1
(f) First we compute
lim−
x→0
1
.
x
It is immediate to see that the limit is divergent to −∞ since the denominator will be negative and very small, while the denominator is constant
equal to 1. Because
lim−
x→0
1
diverges to −∞,
x
1
we conclude that the exponent in e x is negative and very large in absolute
value. The values of the exponential is then getting very close to 0 (recall
that the graph of the exponential tends to 0 as the exponent approaches
−∞). Hence
1
lim e x = 0.
x→0−
(g) If we try to plug in x = 4 in the given limit we obtain the absolute worst
scenario: “ 00 ”, since in this case we can’t even anticipate what is the
answer to our problem!
For this problem the method is to multiply both numerator and denominator by the conjugate of
√
x2 + 9 − 5,
which is
√
x2 + 9 + 5.
4
MATH 100: PROF. DRAGOS GHIOCA
So,
√
x2 +9−5
x−4
√
√
2 +9−5
2
limx→4 xx−4
· √xx2 +9+5
+9+5
limx→4
=
= limx→4
x2 +9−25
√
(x−4)( x2 +9+5)
= limx→4
2
x√
−16
(x−4)( x2 +9+5)
(x−4)(x+4)
√
(x−4)( x2 +9+5)
limx→4 √xx+4
2 +9+5
4+4
√
= 42 +9+5
= 45 .
= limx→4
=
Problem 2.
(a) Since the function f (x) changes its definition from the left to the right of
0, we have to split the above limit into left and right limits. So,
lim f (x) = lim− x = 0
x→0−
x→0
and
lim f (x) = lim+ x2 = 0.
x→0+
x→0
Since both the left and right limits exist and are equal, we conclude that
also
lim f (x) exists and it equals 0.
x→0
(b) Since the function g(x) changes its definition from the left to the right of
1, we have to split the above limit into left and right limits. So,
lim g(x) = lim− 5 − x = 4
x→1−
x→1
and
lim g(x) = lim+ ln(x) = 0.
x→1+
x→1
In this case, the left and right limits exist, but they are not equal; so we
conclude that
lim g(x) does not exist.
x→1
Problem 3.
(a)
change in position
time elapsed
Therefore on the interval of time t = 1 second to t = 5 seconds, the time
elapsed is 5 − 1 = 4 seconds.
On the other hand, the change in position is computed by finding the
position of the particle after 1 seconds, and respectively after 5 seconds
(each time using the displacement equation given in the problem). So,
after t = 1 seconds, the particle is at 3 · 12 + 1 = 4 meters from the origin,
average velocity =
PRACTICE PROBLEMS: SET 1
5
while after t = 5 seconds the particle is at 3 · 52 + 5 = 80 meters from the
origin. In conclusion the average velocity on the interval [1, 5] is
80 − 4
= 19 m/s.
5−1
(b) The instantaneous velocity after t = 2 seconds is computed as the limiting
value of the average velocities over shorter and shorter time periods that
start at t = 2. More precisely, we compute the average velocity of the
particle over the interval of time [2, 2 + h] and compute the limit of these
velocities as h approaches 0. So, we need to compute the following limit:
(3(2+h)2 +(2+h))−(3·22 +2)
h
2
limh→0 (3(4+h +4h)+(2+h))−(12+2)
h
12+3h2 +12h+2+h−14
= limh→0
h
2
= limh→0 3h +13h
h
limh→0
=
= limh→0 3h + 13
= 13.