MATH 100 Winter 2006 midterm exam solutions

Solutions for Math 100 2006 Midterm

1. Each limit exists and is given by

(a) p
sin (x 1) sin (x 1) 1 + 3x + 2
lim p = lim p p
x!1 1 + 3x 2 x!1 1 + 3x 2 1 + 3x + 2


sin (x 1) 4 sin (x 1) 4
= 4 lim = lim = :
(1 + 3x 4)
x!1 3 x!1 x 1 3
p
(b) Use Squeeze Theorem. Since 1 sin ( =x) 1 and x2 = jxj, it
follows that
jxj p jxj p
x2 esin( =x)
e jxj =) lim lim x2 esin( =x)
lim e jxj
e x!0 e x!0 x!0
p p
=) 0 lim x2 esin( =x)
0 =) lim x2 esin( =x)
= 0:
x!0 x!0
p
x2 + 1 x
2. The vertical asymptote of y = is given by x = 2 since the
x+2
denominator is zero at that point but the numerator is not. The horizontal
asymptotes are determined by
r r
p 1 1
jxj 1 + 2 x 1+ 2 1
x2 + 1 x x x
lim = lim = lim = 0;
x!1 x+2 x!1 2 x!1 2
x 1+ 1+
x x
r r
p 1 1
jxj 1 + 2 x 1+ 2 +1
x2 + 1 x x x
lim = lim = lim = 2;
x! 1 x+2 x! 1 2 x! 1 2
x 1+ 1+
x x
so that the horizontal asymptotes are given by y = 0 and y = 2.
3. The derivatives are given by
p 1
(a) y 0 = 3 x + x2 + x + 1 (2x + 1) + ex ;
2x + cos (x)
(b) y 0 = 2;
1 + (x2 + sin (x))
1 + y sin (xy)
(c) 1 + y 0 = sin (xy) (y + xy 0 ) =) y 0 = ;
1 + x sin (xy)
(d) Noting that 10x tan(x) = eln(10)x tan(x) , we …nd

y 0 = ln (10) 10x tan(x) tan (x) + x sec2 (x) :


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