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MAT137Y, 2009–2010 Winter Session, Solutions to Term Test 1

1. Evaluate the following limits. (Do not prove them using the formal deﬁnition of limit.)

(10%) (i) lim

x→4

1

x−1

4

x2−x−12.

Obtaining a common denominator and factoring the denominator, we get

lim

x→4

4−x

4x

(x−4)(x+3)=lim

x→4−1

4x(x+3)=−1

112.

(10%) (ii) lim

x→1

x−√x2−5x+5

√x−1.

We multiply top and bottom by both conjugates, giving us

lim

x→1

x−√x2−5x+5

√x−1·x+√x2−5x+5

x+√x2−5x+5·√x+1

√x+1=lim

x→1

[x2−(x2−5x+5)](√x+1)

(x−1)(x+√x2−5x+5)

=lim

x→1

5(x−1)(√x+1)

(x−1)(x+√x2−5x+5)

=lim

x→1

5(√x+1)

x+√x2−5x+5=5.

2.

(10%) (i) Suppose θis in the ﬁrst quadrant, and cosθ=1

5. Find the exact value of sin 2θ.

We can do this by drawing a triangle. Alternatively, since sin2θ+cos2θ=1, we have

sin2θ=1−cos2θ=1−1

25 =24

25 . Hence, sinθ=√24

5=2√6

5, since sin θ>0 as we are

in the ﬁrst quadrant.

Therefore, sin2θ=2 sinθcos θ=2·2√6

5·1

5=4√6

25 .

(10%) (ii) Consider the function f(x) = |x−3|+|x|. Express f(x)as a piecewise function by

eliminating the absolute values, and (without plotting any points) sketch the graph of

f.

To express fas a piecewise function, we eliminate the absolute value ﬁrst:

|x−3|=(x−3,x≥3

3−x,x<3,|x|=(x,x≥0

−x,x<0.

Therefore,

if x<0,|x−3|+|x|=3−x+ (−x) = 3−2x,

if 0 ≤x<3,|x−3|+|x|=3−x+x=3,

if x≥3,|x−3|+|x|=x−3+x=2x−3.

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