Louisiana State University
Math1550 Test 1: Answer Key
1. The graphs of f and g are given.
Use the above graphs to evaluate each limit, if it exists. (If an answer does not exist, write DNE.)
(a) lim [f (x) + g(x)] = lim f (x) + lim g(x)] = 2 + 0 = 2
x→2
x→2
x→2
(b) lim [f (x) + g(x)] = DN E
x→1
Note that limx→1 f (x) = 1 and limx→1 g(x) = DN E since limx→1− g(x) = 2 and limx→1+ g(x) =
1, therefore the sum of the limits does not exist.
(c) lim [f (x) · g(x)] = lim f (x) · lim g(x) = 0 ·
x→0
x→0
x→0
3
=0
2
f (x)
= DN E
x→−1 g(x)
Note that limx→−1 g(x) = 0 which is in denominator and limx→−1 f (x) = 1 6= 0 then the
quotient limits does not exist.
(d) lim
(e) lim [x3 f (x)] = lim x3 lim f (x) = 23 · 2 = 16
x→2
(f) lim
x→1
x→2
x→2
q
p
√
3 + f (x) = lim 3 + lim f (x) = 3 + 1 = 2
x→1
x→1
From the graph of f and g,
(a) State the sets on which f is continuous
f is continuous whenever x ∈ (−∞, 2) ∪ (2, ∞)
(b) State the sets on which g is discontinuous
g is discontinuous whenever x ∈ {1}
2. Find the limits if they exist or if doesn’t explain why?
t2 − 81
(t − 9)(t + 9)
(t − 9)
18
= lim
= lim
=
2
t→−9 2t + 19t + 9
t→−9 (2t + 1)(t + 9)
t→−9 (2t + 1)
17
(a) lim
8x5 −8x
(b) lim 2
x→1
8·15 −8·1
+x = 2
+ 1 = 20 + 1 = 2
1
3. [10 pts] Use the given graph of f (x) =
to find a number δ such that if |x − 2| < δ then
x
1 1
− < 0.2
x 2
On the left side of x = 2, we need |x − 2| < | 10
− 2| = 47 . On the right side of x = 2, we need
7
− 2| = 43 . For both of these conditions to be satisfied at once, we need the more
|x − 2| < | 10
3
restrictive of the two to hold, that is, |x − 2| < 74 (the minimum of 47 and 43 ). So we can choose
δ = 47 , or any smaller positive number.
4. Find the horizontal and vertical asymptotes of the curve y =
8 + x4
.
x2 (1 − x2 )
Vertical asymptotes happen when:
8 + x4
.
x2 − x4
Note that y =
x2 (1 − x2 ) = 0
thus,
x2 = 0
and
x2 − 1 = 0
x=0
and
x = 1 and x = −1
Horizontal asymptotes happen when:
8 + x4
y = lim 2
= lim
x→∞ x − x4
x→∞
8
x4
1
x2
+1
8 + x4
= lim
= −1, and, y = lim 2
x→−∞
x→−∞ x − x4
−1
8
x4
1
x2
+1
= −1.
−1
Therefore, the vertical asymptotes are: x = 0, x = −1, x = 1 and the vertical asymptote is y = −1
5. Find the equation of the tangent line to the curve y = x2 + x at x = 1.
Note that if x = 1 then y = 12 + 1 = 2. The slope mt for any tangent line to the above curve is
given by mt = y 0 = 2x + 1.
In particular, if x = 1 then the slope is mt (1) = 3 thus, the equation of the line will have the form
y = 3x + b. Using the point (1, 2) we have that b = −1. Therefore, the equation of the tangent
line to the curve y = x2 + x at x = 1 is
y = 3x − 1
√
4 − x, using the definition of derivative.
p
p
√
√
4 − (x + h) − 4 − x
4 − (x + h) + 4 − x
f (x + h) − f (x)
0
f (x) = lim
= lim
·p
√
h→0
h→0
h
h
4 − (x + h) + 4 − x
6. Find the derivative of f (x) =
4−x−h−4+x
4 − (x + h) − (4 − x)
p
= lim p
√
√
h→0 h(
4 − (x + h) + 4 − x) h→0 h( 4 − (x + h) + 4 − x)
−h
−1
= lim p
= lim p
√
√
h→0 h(
h→0
4 − (x + h) + 4 − x)
( 4 − (x + h) + 4 − x)
−1
−1
= √
= p
√
2 4−x
( 4 − (x + 0) + 4 − x)
= lim