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(MATH023)[2011](f)midterm~=vfemuek^_70631.pdf
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Math 023 -Honors Calculus I Midterm Test, Fall Semester, 2011
Time Allowed: 2 Hours Total Marks: 100
1. (40 Marks) Find the following limits: n + sin n
(a) lim ;
n n2 +1
n)1/n2
(b) lim (1+2n + + n ;
n
n
x+ x2n . 2
(c) lim , where n is a positive integer;
x1 x . 1
..1/x
1 + sin x
(d) lim .
x0 1 . sin x
2. (20 Marks) Show that the equation sin x = cos2 x has only one solution on
0, .
2
3.
(20 Marks) Use the .- language to show that x.1/2 is continuous at x = a for any a> 0.
4.
(20 Marks) Suppose lim an = a. Show that the sequence {xn},
n
xn =(a1 . a)1 +(a2 . a)2 + +(an . a)n ,
is convergent.
1
Solutions
1. (40 Marks) Find the following limits: n + sin n
(a) lim ;
n n2 +1 n 1
Solution It is clear that lim = lim = 0. We
n n2 +1 n n(1 + 1/n2)
sin n
claim that lim = 0. In fact, for any .> 0, if n> 1/., then
n n2 +1
sin n
n
2 +1 11
< <..
2 +1
n
n
Hence,
n + sin nn sin n
lim = lim + lim =0+0=0.
n n2 +1 n n2 +1 n n2 +1
n)1/n2
(b) lim (1+2n + + n ;
n
Solution Since
n)1/n2 n)1/n2 2)1/n2 1/n
1 < (1 + 2n + + n< (n n (n n,
1/n 2)1/n2
and lim n = lim (n = 1, by the Sandwich Rule, we have
n n
n)1/n2
lim (1+2n + + n = 1.
n
n
x+ x2n . 2
(c) lim , where n is a positive integer;
x1 x . 1
Solution
xn + x2n . 2
lim
x1 x . 1
(xn . 1) + (x2n . 1)
= lim
x1 x . 1 n.1 n.22n.12n.2
(x . 1)(x+ x+ + 1) + (x . 1)(x+ x+ + 1) = lim
x . 1
x1
n.1 n.22n.12n.2
lim (x
+ 1) + (x
+
+ 1)
+ x
+
+ x
=
x1
= n +2n =3n.
2
..1/x
1 + sin x
(d) lim .
x0 1 . sin x
Solution Since
1 + sin x 2 sin x
=1+ ,
1 . sin x 1 . sin x 2 sin x 2 sin x
with lim = 0, if we put y = , we have
x0 1 . sin x 1 . sin x
1 . sin x 2 sin x 1
lim log 1 + = lim log(1+y) = lim log(1+y)1/y =1.
x0 2 sin x 1 . sin x y0 y y0
Hence,
..1/x
1 + sin x 1 log(1+sin x )
lim = lim ex 1.sin x
x0 1 . sin x x0
1 2 sin x 1.sin x log(1+ 2 sin x
)
= lim x 1.sin x 2 sin x 1.sin x
e
x0 1 2 sin x 1.sin x log(1+ 2 sin x
limx0 )
x 1.sin x 2 sin x 1.sin x
= e
2 sin x 11.sin x log(1+ 2 sin x
limx0 limx0 limx0 )
= e x 1.sin x 2 sin x 1.sin x
2112
= e = e.
2. (20 Marks) Show that the equation sin x = cos2 x has only one solution on
0, .
2
2
Solution Let f(x) = sin x.cosx. It is clear that f is a continuous function
on 0, , and f(0) = .1 < 0, f(/2) = 1 > 0. Thus, by the Intermediate
2 ..
Value Theorem, we know that f has at least one root on 0, .
2 We prove that f has at most one solution by proving that it is strictly
increasing. In fact, for any x, y 0, , with x<y, we have
2
22
f(x) . f(y) = sin x . cos x . sin y . cos y
= sin x . 1 + sin2 x . sin y . 1 + sin