This can be achieved by making f (3) unequal to
the limit, but make sure that the function is defined
at x = 3. See section 2.5 of the text.
First Test Answers
Math 120 Calculus I
3. [10; 5 points each property] On asymptotes.
a. Sketch the graph of a function f such that
September, 2013
lim f (x) = 0 and lim+ f (x) = −∞.
x→2−
Scale. 90–100 A, 80–89 B, 65–79 C. Median 80.
x→2
The graph of the function should should be
asymptotic to the vertical line x = 2. See section
1. [12] On limits of average rates of change. Let 2.6 of the text.
f (x) = 5x2 + 4.
b. Sketch the graph of a function f such that
lim
f (x) = 1.
a. [4] Write down an expression that gives the
x→−∞
average rate of change of this function over the inThe graph of the function should should be
terval between x and x + h, and simplifly the exasymptotic to the horizontal line y = 1. See section
pression.
2.6 of the text.
f (x + h) − f (x)
4. [28; 7 points each part] Evaluate the following
h
2
2
(5(x + h) + 4) − (5x + 4)
limits. If a limit diverges to ±∞ it is enough to say
=
that it doesn’t exist.
h
x2 − 4x + 4
b. [8] Compute the limit as h → 0 of that average
a. lim 2
x→1 x − x − 2
rate of change.
As x approaches 1, the numerator approaches 1
2
2
while the denominator approaches −2, so the quo(5(x + h) + 4) − (5x + 4)
lim
tient approaches − 12 .
h→0
h
x2 − 4x + 4
5x2 + 10xh + 5h2 + 4 − 5x2 − 4)
b.
lim
= lim
x→2 x2 − x − 2
h→0
h
The expression needs to be simplified before tak10xh + 5h2
= lim
ing the limit.
h→0
h
= lim (10x + 5h) = 10x
(x + 2)(x − 2)
x+2
h→0
lim
= lim
= 43
x→2 (x + 1)(x − 2)
x→2 x + 1
2. [10; 5 points each] On the intutitive concept of
3x2 − 2x + 1
limit and continuity.
c. lim
x→∞
9x3 + x
a. [5] Sketch the graph y = f (x) of a function
The numerator has a lower degree than the defor which lim f (x) does not exist.
nominator, so as x → ∞, the limit approaches 0.
x→3
This
can be seen by dividing the numerator and
There are many such graphs. For example, if
3
there’s a jump in the value of f at x = 3, then that denominator by x
limit won’t exist. See section 2.4 of the text.
3x2 − 2x + 1
3/x − 2/x2 + 1/x3
lim
=
lim
b. [5] Sketch the graph y = f (x) of a function x→∞
x→∞
9x3 + x
9 + 1/x2
defined everywhere, the limit lim f (x) does exist,
0−0+0
x→3
=
=0
but f is not continuous at x = 3.
9+0
1
5x
.
8 sin x
Since f (x) approaches 4, and g(x) approaches 9,
the numerator approaches 9 while the denominator
sin x
x
approaches −1. Therefore quotient approaches −9.
Recall that lim
= 1. Therefore lim
=
x
x→0
x→0
x
sin x
b. lim
1, hence this limit equals 58 .
x→π/2 g(x) − f (x) − sin x
d. lim
x→0
As x approaches π/2, sin x approaches 1. Therefore the denominator approaches 0. But the nuConsider the limit lim (9 − 2x) which, of course, merator approaches π/2, so the limit doesn’t exist.
x→3
p
has the value 3. Since it has the value 3, that means
(g(x))2 + (f (x))2
c. lim
x→π/2
that for each > 0, there exists some δ > 0, such
that for all x, if 0 < |x−3| < δ, then |(9−2x)−3| <
The sum (g(x))2 + (f (x))2 approaches 25. Since
.
the square root function is continuous, the limit
1
Let = 3 . Find a value of δ that works for this approaches the square root of 25, which is 5.
.
5. [15] On the formal definition of limit.
You need to find a value of δ so that
0 < |x − 3| < δ implies |(9 − 2x) − 3| < 31 .
The expression |(9−2x)−3| can be rewritten as |6−
2x| which equals 2|x − 3|. Therefore, the condition
|(9 − 2x) − 3| < 13 is equivalent to |x − 3| < 16 . Thus,
you need to find a value of δ so that
0 < |x − 3| < δ implies |x − 3| < 16 .
Such a value is δ = 61 .
6. [10] Suppose that θ √
is an angle between π/2
and π, and that sin θ = 21 2. Determine the value
of cos θ.
√
Since sin θ = 12 2, the Pythagorean identity
sin2 θ + cos2 θ = 1 implies 12 + cos2 θ = 1. Hence,
√
cos2 θ = 12 , so cos θ = ± 12 2. Since θ is an angle between π/2 √
and π, the cosine of θ is negative.
Thus cos θ = − 12 2.
7. [15; 5 points each part] Suppose that
lim f (x) = 4 and lim g(x) = 5. Evaluate each
x→π/2
x→π/2
of the following limits, or explain why it doesn’t
exist
f (x) + g(x)
a. lim
x→π/2 f (x) − g(x)
2