Quiz 7
12pm
Fall 2012
Solutions
Math 0220
1. Find the limit. Use l’Hospital rule if appropriate.
ln t
.
sin πt
ln t 0 H
1/t DSP
1
1
lim
” 0 ” = lim
=
=− .
t→1 sin πt
t→1 π cos πt
π(−1)
π
(a) [2 points] lim
t→1
Solution:
(b) [3 points] lim x tan(1/x).
x→∞
Solution:
lim x tan(1/x) [”0 · ∞”] = lim
x→∞
x→∞
tan(x−1 ) 0 H
”0” =
x−1
lim
x→∞
sec2 (x−1 )(−x−2 )
−x−2
= lim sec2 (x−1 ) = sec2 0 = 1.
x→∞
x
satisfies the hypotheses of the MVT on the
x+1
interval [0, 3]. Then find all numbers c that satisfy the conclusion of the MVT.
2. [5 points] Verify that the function f (x) =
Solution: The only discontinuity that f (x) has is at −1. Hence f (x) is continuous on [0, 3]. It
is a rational function and hence differentiable on (0, 3). Therefore, f (x) satisfies the hypotheses
of the MVT on the interval [0, 3].
By the MVT there is a number c such that f 0 (c) =
f 0 (x) =
f (3) − f (0)
3/4 − 0
1
=
= .
3−0
3
4
1
x+1−x
=
.
(x + 1)2
(x + 1)2
Therefore, to find c we need to solve the equation
1
1
=
⇔ (c + 1)2 = 4.
2
(c + 1)
4
Then c + 1 = 2 or c + 1 = −2. Two solutions c = 1 and c = −3. The last is not in (0, 3).
Hence, c = 1.
1
bonus problem [5 points extra] Evaluate the limit lim 3(sin x)tan x .
x→0
Solution:
tan x
(sin x)tan x = eln(sin x)
= etan x·ln(sin x)
lim tan x·ln(sin x)
lim 3(sin x)tan x = 3 lim (sin x)tan x = 3 lim etan x·ln(sin x) = 3e x→0
x→0
x→0
x→0
lim tan x·ln(sin x) [”0·∞”] = lim
x→0
x→0
cot x
ln(sin x) ∞ H
DSP
” ∞ ” = lim
= − lim cos x sin x = 0.
2
x→0 − csc x
x→0
cot x
Hence, lim 3(sin x)tan x = 3e0 = 3.
x→0
2