MATH 126, QUIZ 1
Instructions:
Solve the following problems in a piece of paper with your name on it.
(1) Determine the value of the limit
1
lim (1 − 2x) x .
x→0
Solution
We have
1
ln((1 − 2x) x ) =
ln((1 − 2x))
,
x
and hence, by Lôpital's rule,
ln((1 − 2x))
−2
= lim
= −2.
x→0
x→0 1 − 2x
x
1
lim ln((1 − 2x) x ) = lim
x→0
Therefore, taking exponential in both sides, and using the continuity of the exponential function, we get
1
1
1
lim (1 − 2x) x = lim eln((1−2x) x ) = elimx→0 ln((1−2x) x ) = e−2 ,
x→0
x→0
where the last equality follows by an application of Lôpital's rule,
indenite integral
(2) Determine the value of the
Z
Solution
Making the substitution
Z
sin(x)
dx =
2 + cos(x)
Z
sin(x)
dx.
2 + cos(x)
u = 2 + cos(x), du = − sin(x)dx,
we obtain
−u−1 du = − ln(u) + C = − ln(2 + cos(x)) + C.
(3) Determine the value of the following
denite integrals
(a)
π
2
Z
cos(x) sin(sin(x))dx.
0
(b)
Z
0
Solution
π
2
Z
Date
we obtain
1
sin(u)du = − cos(1) − (− cos(0)) = − cos(1).
cos(x) sin(sin(x))dx =
0
ez + 1
dz.
ez + z
u = sin(x), du = cos(x)dx,
(a) Making the substitution
Z
1
0
: August 31.
1
2
MATH 126, QUIZ 1
(b) Making the substitution
Z
0
1
z
e +1
dz =
ez + z
Z
1
u = ez + z , du = ez + 1,
e+1
we get
1
du = ln(e + 1) − ln(1) = ln(e + 1).
u