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Unformatted text preview: Solutions to Bonus Quiz A
www.math.uﬂ.edu/˜harringt
February 17, 2008
1. Find all the vertical asymptotes for f (x) =
Note that f (x) = (x−1) sin(x)
x3 − x = x− 1
x− 1 ∗ sin x
x ∗ (x−1) sin(x)
.
x3 − x 1
.
x+1 So we have holes at x = 1 and at x = 0 since limx→0
limx→1 x−1 = 1.
x− 1 sin x
x = 1 and Thus, we have ONE vertical asymptote at x = −1.
2. Solve for x: e3x = 5e4x+2
Note: There are multiple ways to do this problem. I will demonstrate
only two. e3x =
e3x − 5e4x+2 =
e3x − 5e3x+x+2 =
e3x (1 − 5ex+2 ) =
1 − 5ex+2 =
1=
1/5 =
ln 1/5 − 2 = 5e4x+2
0
0
0 Note: that e3x = 0 for all x
0 We divided both sides by e3x
5ex+2
ex+2
x Or....
1 e3x
ln(e3 x)
3x
3x
−x
x 5e4x+2
ln(5 ∗ e4x+2 )
ln(5) + ln(e4x+2 )
ln(5) + 4x + 2
ln(5) + 2
− ln(5) − 2 =
=
=
=
=
= 3. Evaluate the following limit: limx→0 tan x
x tan x
sin(x)
sin(x)
1
= lim
= lim
∗ lim
=1∗1=1
x→0
x→0 x cos(x)
x→0
x→0 cos x
x
x
lim 4. Determine which of the following is even: x sin x or
answer. x
.
x Explain your Recall that a function is even when f (−x) = f (x).
(−x) sin(−x) = −x ∗ (− sin(x)) = x ∗ sin x Thus, x sin x is EVEN.
−x
−x = x
−x = − x Hence,
x x
x is ODD. 5. Compute the following limits. Let
3 x + 1 for x < 0
x2 − x for 0 ≤ x < 1
f (x) =
√
x + x for x > 1
(a) limx→0 f (x) =Does Not Exist
Since limx→0+ f (x) = 02 + 0 = 0 and limx→0− f (x) = 03 + 1 = 1
(b) limx→1− f (x) = 12 + 1 = 2
(c) limx→0+ f (x) = 02 + 0 = 0
(d) limx→1 f (x) = 2
√
Since limx→1+ f (x) = 1 + 1 = 2 and limx→1− f (x) = 12 + 1 = 2 2 ...
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This note was uploaded on 07/10/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
 Fall '08
 ALL
 Calculus, Geometry, Asymptotes

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