2004 Test 1 solution

25 views3 pages
10 Apr 2012
School
Department
Course
Professor
MAT 137Y 2004-2005 Solutions to Term Test 1
1. Evaluate the following limits, or show that the limit does not exist.
(10%) (i) lim
x0
1+x1x
x.
lim
x0
1+x1x
x=lim
x0
1+x1x
x1+x+1x
1+x+1x=lim
x0
(1+x)(1x)
x(1+x+1x)
=lim
x0
2x
x(1+x+1x)=lim
x0
2
1+x+1x=1.
(10%) (ii) lim
x0
(tanx)(tan2x)
xtan3x.
lim
x0
(tanx)(tan2x)
xtan3x=lim
x0
sinx
x·sin2x
2x
3x
sin3x
2x
3x
cos3x
(cosx)(cos2x)=1·1·1·2
3·1
(1)(1)=2
3.
(10%) (iii) lim
x4|4x|
x2x12.
By definition, |4x|=(x4,x4,
4x,x<4.Taking cases on x, we have
lim
x4+|4x|
x2x12 =lim
x4+
x4
(x4)(x+3)=lim
x4+
1
x+3=1
7.
Similarly,
lim
x4|4x|
x2x12 =lim
x4
4x
(x4)(x+3)=lim
x41
x+3=1
7.
Since lim
x4|4x|
x2x12 6=lim
x4+|4x|
x2x12, it follows the limit does not exist.
2.
(8%) (i) Solve the inequality x29
3x+1<1.
Moving all terms to one side, we solve
x29
3x+11<0x29(3x+1)
3x+1<0x23x10
3x+1<0(x5)(x+2)
3x+1<0.
Drawing a chart of signs, we get
x<22<x<1
31
3<x<5x>5
x5 +
x+2+ + +
3x+1− − + +
(x5)(x+2)
3x+1++
Hence the solution is x(,2)(1
3,5).
1
Unlock document

This preview shows page 1 of the document.
Unlock all 3 pages and 3 million more documents.

Already have an account? Log in

Get OneClass Grade+

Unlimited access to all notes and study guides.

Grade+All Inclusive
$10 USD/m
You will be charged $120 USD upfront and auto renewed at the end of each cycle. You may cancel anytime under Payment Settings. For more information, see our Terms and Privacy.
Payments are encrypted using 256-bit SSL. Powered by Stripe.