1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
x3 − 5x + 7
x3 − 5x + 7
=
lim
∗
x→∞ x3 + 6x4 + 2
x→∞ x3 + 6x4 + 2
1. lim
2. lim
p
x2
x→∞
lim √
x→∞
3. lim
x→∞
5x + 4
∗
2
x + 5x + 4 + x
2x2
lim √
x→∞
x2 + 5x + 4
∗
2x2 + 5x + 4 + x
lim 1 +
= lim
5+
√
x→−∞
r
lim 1 −
x→−∞
0
=0
6
= lim q
x2 + 5x + 4 + 1 x→∞ 1 +
5+
5
x
4
x
+
4
x2
=
+1
5
2
√
2x2 + 5x + 4 + x
2x2 + 5x + 4 − x2
+ 5x + 4 − x) ∗ √
= lim √
=
2x2 + 5x + 4 + x x→∞ 2x2 + 5x + 4 + x
x→∞ √1
x2
√
2x + 5 + x4
= lim q
=∞
2x2 + 5x + 4 + 1 x→∞ 2 + 5 + 42 + 1
x
x
x+5+
1
x2
1
x2
4
x
1+
= lim
−1 √
−1
√
x2 x2
x→−∞
√
2x4 + 3
1
√
3
=1+ 2
4
x
2x2 + 3
x+
= lim
x→−∞
x
2+
=
4
x
√
2x4 + 3
x2 + 2x4 + 3
= lim
∗
x→−∞
x2
x2
2+
x→−∞
lim
1
x
1
x
2x2
√
√
x→−∞
x+
x→∞ √1
x2
x→∞
1
5
7
x − x3 + x4
x→∞ 1 + 6 + 24
x
x
= lim
√
x2 + 5x + 4 + x
x2 + 5x + 4 − x2
+ 5x + 4 − x) ∗ √
= lim √
=
x2 + 5x + 4 + x x→∞ x2 + 5x + 4 + x
= lim
+ 5x + 4 − x = lim (
r
5.
1
x
1
x
p
x2 +
lim
x2
+ 5x + 4 − x = lim (
p
x→∞
4.
p
1
x4
1
x4
September 17
√
2x2 + 3
∗
x
1
x
1
x
1+
= lim
−1
√
x2
x→−∞
√
2x2 + 3
1
=
√
3
=1− 2
2
x
6. Is the function, f (x), continuous?
(a) f (x) =

2x










for x < 1
6x2 − 4 for 1 ≤ x ≤ 2
for x > 2
4x
Since each piece is a polynomial, then the pieces are continuous.
f (x) is continuous at x = 1 since:
lim f (x) = lim 2x = 2 = lim 6x2 − 4 = lim f (x).
x→1−
x→1−
x→1+
x→1+
f (x) is not continuous at x = 2 since:
lim f (x) = lim 6x2 − 4 = 20 6= 8 = lim 4x = lim f (x).
x→2−
(b) f (x) =
x→2−

x+7




x−5
x3 − 5x + 2
x→2+
x→2+
for x < 1
for x > 1
Both pieces of this of this function are continuous for the defined sections. The top piece
does have a vertical asymptote at x = 5, but the top piece is only used for x < 1. Notice
that the lim f (x) = −2, however, the function is NOT defined at x = 1 so it is not
x→1
continuous at x = 1.
2
151 WebCalc Fall 2002-copyright Joe Kahlig
7. Find the values of A and B such that the function f (x) will be continuous.
f (x) =
For f (x) to be continuous, we need

x
Ax + B
 2
5x
for x ≤ −1
for −1 < x ≤ 2
for x > 2
lim f (x) = lim f (x) and lim f (x) = lim f (x).
x→−1−
x→−1+
lim f (x) = lim + f (x)
x→−1−
x→−1
lim − x = lim + Ax + B
x→−1
x→−1
−1 = −A + B
lim f (x) = lim f (x)
x→2−
x→2+
lim Ax + B = lim 5x2
x→2−
2A + B = 20
x→2+
Now solve the system of equations to get A = 7 and B = 6.
x→2−
x→2+