Solutions for Math 100 2016 Midterm
1. Each limit exists and is given by
(a) p p
p x2 + x + x x2 + x x
lim x2 + x + x = lim p
x! 1 x! 1 x2 + x x
x2 + x x2 1 1
= lim p = lim p = :
x! 1 jxj 1 + 1=x x x! 1 1 + 1=x + 1 2
(b)
tan (4x) sin (4x) 2 sin (2x) cos (2x)
lim = lim = lim = 2:
x!0 sin (2x) x!0 sin (2x) cos (4x) x!0 sin (2x)
2. The ellipse crosses the x-axis when y = 0, i.e.,
x2 + 0 + 0 = 4 =) x = 2.
Hence the points where the ellipse crosses the x-axis are given by ( 2; 0)
and (2; 0). To show that the tangent lines at these points are parallel
means to show that the slopes of the two tangent lines is the same and
that means to show that y 0 ( 2) = y (2), thinking of y (x). From the
equation of the ellipse, using implicit di¤erentiation, we have
2x + y + xy 0 + 2yy 0 = 0;
so that the point ( 2; 0) we …nd
4+0 2y 0 + 0 = 0 =) y 0 = 2;
and at the point (2; 0) we …nd
4 + 0 + 2y 0 + 0 = 0 =) y 0 = 2;
which are the same. Hence, we conclude that the tangent lines are parallel
at the points ( 2; 0) and (2; 0) on the ellipse.
3. The derivatives are given by
cos (earctan x ) earctan x
(a) y 0 = ;
1 + x2
2x
(b) y 0 = 2 ;
x +1
y + xy 0
(c) q = sin (x + y) (1 + y 0 ) ;
2
1 (xy)
(d) y 0 = x 1
+ ln ( ) x
:
1
1. Each limit exists and is given by
(a) p p
p x2 + x + x x2 + x x
lim x2 + x + x = lim p
x! 1 x! 1 x2 + x x
x2 + x x2 1 1
= lim p = lim p = :
x! 1 jxj 1 + 1=x x x! 1 1 + 1=x + 1 2
(b)
tan (4x) sin (4x) 2 sin (2x) cos (2x)
lim = lim = lim = 2:
x!0 sin (2x) x!0 sin (2x) cos (4x) x!0 sin (2x)
2. The ellipse crosses the x-axis when y = 0, i.e.,
x2 + 0 + 0 = 4 =) x = 2.
Hence the points where the ellipse crosses the x-axis are given by ( 2; 0)
and (2; 0). To show that the tangent lines at these points are parallel
means to show that the slopes of the two tangent lines is the same and
that means to show that y 0 ( 2) = y (2), thinking of y (x). From the
equation of the ellipse, using implicit di¤erentiation, we have
2x + y + xy 0 + 2yy 0 = 0;
so that the point ( 2; 0) we …nd
4+0 2y 0 + 0 = 0 =) y 0 = 2;
and at the point (2; 0) we …nd
4 + 0 + 2y 0 + 0 = 0 =) y 0 = 2;
which are the same. Hence, we conclude that the tangent lines are parallel
at the points ( 2; 0) and (2; 0) on the ellipse.
3. The derivatives are given by
cos (earctan x ) earctan x
(a) y 0 = ;
1 + x2
2x
(b) y 0 = 2 ;
x +1
y + xy 0
(c) q = sin (x + y) (1 + y 0 ) ;
2
1 (xy)
(d) y 0 = x 1
+ ln ( ) x
:
1
