MATH 100 Winter 2011 midterm exam solutions

Solutions for Math 100 2011 Midterm

1. Each limit exists and is given by

(a)
2 3
p p
( x 1)
3
( x 1)
3

lim sin = lim sin 4 p p p
5
x!1 x 1 x!1
(3x 1)
3
x2 + 3 x + 1
! p
3
= lim sin p
3 p = sin = :
x!1 x2 + 3
x+1 3 2
where we have used
p
a3 b3 = (a b) a2 + ab + b2 with a = 3
x and b = 1;

in the denominator.
(b)
r r !
p 2 2
lim x+ x2 + 2x = lim x+jxj 1 + = lim x 1 1+
x! 1 x! 1 x x! 1 x
q q
2 2
x 1 1+ x 1+ 1+ x x 1 1 x2
= lim q = lim q = 1:
x! 1 2 x! 1
1+ 1+ x 1 + 1 + x1

2. For the line y = 3x + b to be a tangent line to the curve y = 1 3x + 2ex ,
we require that
d (1 3x + 2ex )
3= = 3 + 2ex =) x = ln (3) ;
dx
and that
(3x + b)jx = ln(3) = (1 3x + 2ex )jx = ln(3)
() 3 ln (3) + b = 1 3 ln (3) + 6 =) b = 7 6 ln (3) :

3. The derivatives are given by

2x 1 + x4 4x3 1 + x2 2x 1 2x2 x4
(a) y 0 = 2 = 2 ;
(1 + x4 ) (1 + x4 )
(b) y 0 = 1;
(1 + y 0 ) cos (x + y)
(c) sin (y) y 0 = ;
2 + sin (x + y)
y + xy 0
(d) ln (2) 2y y 0 = 2:
1 + (1 + xy)

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