2010 Test 1 solution

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10 Apr 2012
School
Department
Course
Professor
MAT137Y, 2010-2011 Winter Session, Solutions to Term Test 1
1. Evaluate the limits which exist. (Do not prove them using the formal definition of limit.
You may not use L’Hopital’s Rule.)
(a) (8%) lim
x0
tan(3x2) + sin2(5x)
x2
Note that
lim
x0
tan(3x2)
x2= lim
x0
sin(3x2)
x2
1
cos(3x2)= lim
x0
sin(3x2)
3x2
3
cos(3x2)= (1) 3
1= 3,
and
lim
x0
sin2(5x)
x2= lim
x0sin(5x)
x2
= lim
x0sin(5x)
5x2
·25 = (12)(25) = 25.
Thus
lim
x0
tan(3x2) + sin2(5x)
x2= lim
x0
tan(3x2)
x2+ lim
x0
sin2(5x)
x2= 3 + 28 = 28.
(b) (5%) lim
x3+
x3
|x3|
Since x > 3 in the limit, we see that |x3|=x3, so
lim
x3+
x3
|x3|= lim
x3+
x3
x3= lim
x3+
1
x3,
which does not exist.
(c) (8%) lim
x1
x2x
x1
Multiplying numerator and denominator by x2+x, we see that
lim
x1
x2x
x1= lim
x1
(x2x)
(x1)
(x2+x)
(x2+x)= lim
x1
x4x
(x1)(x2+x)
= lim
x1
x(x31)
(x1)(x2+x)= lim
x1
x(x1)(x2+x+ 1)
(x1)(x2+x)
= lim
x1
x(x2+x+ 1)
(x2+x)=3
2.
2.
(a) (8%) Solve for θ[0,2π).
2 cos2(2θ)1=0
From the equation we immediately see that
cos2(2θ) = 1
2=cos(2θ) = ±r1
2=±2
2.
1
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