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
x→0
tan(3x2) + sin2(5x)
x2
Note that
lim
x→0
tan(3x2)
x2= lim
x→0
sin(3x2)
x2
1
cos(3x2)= lim
x→0
sin(3x2)
3x2
3
cos(3x2)= (1) 3
1= 3,
and
lim
x→0
sin2(5x)
x2= lim
x→0sin(5x)
x2
= lim
x→0sin(5x)
5x2
·25 = (12)(25) = 25.
Thus
lim
x→0
tan(3x2) + sin2(5x)
x2= lim
x→0
tan(3x2)
x2+ lim
x→0
sin2(5x)
x2= 3 + 28 = 28.
(b) (5%) lim
x→3+
√x−3
|x−3|
Since x > 3 in the limit, we see that |x−3|=x−3, so
lim
x→3+
√x−3
|x−3|= lim
x→3+
√x−3
x−3= lim
x→3+
1
√x−3,
which does not exist.
(c) (8%) lim
x→1
x2−√x
x−1
Multiplying numerator and denominator by x2+√x, we see that
lim
x→1
x2−√x
x−1= lim
x→1
(x2−√x)
(x−1)
(x2+√x)
(x2+√x)= lim
x→1
x4−x
(x−1)(x2+√x)
= lim
x→1
x(x3−1)
(x−1)(x2+√x)= lim
x→1
x(x−1)(x2+x+ 1)
(x−1)(x2+√x)
= lim
x→1
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