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.

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