Math 20A
Final Examination.
December 9, 2008
1.
Find the following limits:
(a)
(2 points)
0
51
lim
x
x
x
→
−
,
By L’Hopital’s Rule,
00
5
1
ln(5)5
0
lim
lim
ln(5)
1
xx
H
x
→→
−−
==
.
(b)
(2 points)
0
lim
log
x
x
x
+
→
,
0
0
2
1
log
lim log
lim
lim
lim
0
11
H
x
x
x
x
x
++
+
+
→
→
===
=
−
−
=
.
(c)
(2 points)
5
0
lim
x
x
x
+
→
.
If
y
=
x
5
x
, then ln(
y
) = 5
x
ln(
x
), and
lim 5 ln( )
5 lim
ln( )
0
=
=
, from (b).
Since
e
x
is a continuous, onetoone function, we have that
.
50
0
lim
1
x
x
xe
+
→
2.
Compute the following integrals:
(a) (3 points)
9
0
xdx
∫
9
99
1/2
3/2
0
22
2
(9)
(0)
18
33
3
xdx
x dx
x
⎡⎤
=−=
⎢⎥
⎣⎦
∫∫
.
(b) (3 points)
5
2
0
25
x dx
−
∫
(Hint: Use geometry and interpret as an area.)
The function
2
25
x
−
represents the top half of a circle with radius 5. The
bounds on the integral from 0 to 5 represent the area in the first quadrant. So,
the integral equals onefourth the area of a circle with radius 5. That is given
by (1/4)
p
(5)
2
= (25/4)
p
.
1
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(6 points) Find the point(s) on the ellipse given by
x
2
+ 4
y
2
= 1 where the tangent
line has slope 1.
Using implicit differentiation, we have:
2
28
0
84
x
x
xy
y
y
y
y
′′
+=
⇒
=
−
=
−
. Setting
1
y
′ =
, we have:
14
4
x
y
y
′
=
=−
⇒ =−
.
Substituting this into our equation for the ellipse, we have:
22
2
1
(4) 4
1 2
0
1
20
yy
y
y
−+=
⇒
=
⇒
=
±
. Plugging these yvalues into our
relationship,
x
= 4
y
, we obtain the two points:
11
4,
20
20
⎛⎞
−
⎜⎟
⎝⎠
and
20
20
−
.
4.
Let
3
()
3
x
fx
x
+
=
−
.
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 Fall '08
 Eggers
 Math, Calculus, Limits, Optimization, local maximum

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