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Math 111
Quiz #3 Solutions
Due Sept. 12
1. Give a detailed explanation of why one of the following equations is true while the other equation
is false.
x2 − 9
x+3
x2 − 9
x+3
(i) lim 2
= lim
(ii) 2
=
x→3 x − 2x − 3
x→3 x + 1
x − 2x − 3
x+1
Equation (i) is true and equation (ii) is false.
In equation (ii), the function on the left of the equals sign has domain all real numbers except
for x = 3 and x = −1. The function on the right side of the equal sign has domain all real numbers
except x = −1. Since the domains of the function on the left is not the same as the domain of the
function on the right, the two functions can not be equal. Another way of saying this is to observe
that the function on the left is undefined at x = 3 while the function on the right is defined at
x = 3, making the two functions different.
In equation (i), we are interested in taking the limit of the same functions that show up in
equation (ii). When we take limits, we don’t care about what happens when x = a (in this case
a = 3). Since, for x-values around 3 but not equal to 3, x − 3 does not equal 0, we can cancel the
(x − 3) term from the numerator and denominator. By doing this we are saying that around x = 3
2 −9
but not at x = 3, x2x−2x−3
does in fact equal x+3
. Hence the two limits in (i) are equal.
x+1
2. Calculate the following limits, if they exist:
√
(a) lim
x→7
x+2−3
x−7
√
lim
x→7
(b) lim+
x→3
√
x+2−3
x+2+3
x+2−9
√
·√
= lim
x→7
x−7
x+2+3
(x − 7)( x + 2 + 3)
x−7
√
= lim
x→7 (x − 7)( x + 2 + 3)
1
= lim √
x→7
x+2+3
1
1
=√
=
6
7+2+3
x2 + x
3−x
This limit is of the form ‘non-zero number divided by zero’ if we try to plug in x = 3. This
is the case where we use words.
As x approaches 3 from the right, the numerator approaches 12, a non-zero constant, while
the denominator becomes smaller and smaller (as small as we want) and is negative. This means
the quotient gets larger and larger (as large as we want) and is negative, hence
lim+
x→3
x2 + x
= −∞.
3−x
(c) lim (e−2x cos x)
x→∞
We can’t use limit law #4 since limx→∞ cos x DNE. We must use the Squeeze Theorem.
−1 ≤ cos x ≤ 1
⇒
−e−2x ≤ e−2x cos x ≤ e−2x
Since limx→∞ −e−2x = 0 = limx→∞ e−2x (use a simple substitution if you want), we have
lim e−2x cos x = 0 by the Squeeze Theorem.
x→∞
3. True or False: Justify your answer if true, give a counterexample if false.
(a) If f (1) > 0 and f (3) < 0, then there exists a number c between 1 and 3 such that f (c) = 0.
False. f must also be continuous on the interval [1, 3] for the statement to be true. Here are
examples of two different functions which satisfies the ‘if’ part of (a) but not the ‘then’ part.
3
2
1
1
2
3
4
1
2
3
4
-1
-2
3
2
1
-1
-2
(b) If p and q are two polynomials, then limx→a (p(x) − q(x)) = p(a) − q(a).
True. Since p(x) and q(x) are polynomials, they are continuous everywhere. By theorem
#4 in section 2.4, p(x) − q(x) is also continuous everywhere. By the definition of continuity, when
taking a limit, we can just plug in x = a. So
lim (p(x) − q(x)) = p(a) − q(a).
x→a
You can also explain this using the substitution property and limit laws from section 2.3.
(c) If limx→∞ f (x) = ∞ and limx→∞ g(x) = ∞, then limx→∞ (f (x) − g(x)) = 0.
False. Infinity is not a number and we can not treat it like it is. Here are examples of
functions f and g which satisfy the ‘if’ part of (c) but not the ‘then’ part.
f (x) = ex :
g(x) = x7 :
lim f (x) = lim ex = ∞
x→∞
x→∞
lim g(x) = lim x7 = ∞,
x→∞
x→∞
but
lim (f (x) − g(x)) = lim (ex − x7 ) = ∞
x→∞
x
x→∞
since e increases more quickly than every polynomial.