Part II: The Answers
200
Now you want a 5 in the denominator of the first fraction and a 9 in the numerator of
the second fraction. Multiplying by 5
5
and also by 9
9
gives you
=
( ) ⋅
( )


 


  ( )








=
( )
=
→
lim
sin
sin
( )( )
x
x
x
x
x
0
5
5
9
9
5
9
1 1 5
9
5 5
9
203.
7
3
Begin by rewriting tan(7x) to get
lim
tan( )
sin( )
lim
sin( )
cos( )
sin( )
lim
sin
x
x
x
x
x
x
x
x
→
→
→
=
=
0
0
0
7
3
7
7
3
( ( )
cos( ) sin( )
lim
sin( )
cos( ) sin(
7
7
1
3
7
1
1
7
1
3
0
x
x
x
x
x
x
⋅






=
⋅
⋅
→
x x )






Now you want 7x in the denominator of the first fraction and 3x in the numerator of the
third fraction. Therefore, multiply by x
x
, 3
3
, and 7
7
to get
lim
sin( )
cos( ) sin( )
( ) cos
( )
x
x
x
x
x
x
→
⋅
⋅
⋅






=
( )
0
7
7
1
7
3
3
7
3
1
1
0
1 7
3 3
7
3
( )
=
204.
8
Begin by breaking up the fraction as
lim
sin ( )
lim
sin( ) sin( ) sin( )
x
x
x
x
x
x
x
x
x
x
→
→
=
⋅
⋅






0
3
3
0
2
2
2
2
Next, you want each fraction to have a denominator of 2x, so multiply by 2
2
, 2
2
, and 2
2
—
or equivalently, by 8
8
— to get
lim
sin( ) sin( ) sin( )
( )( )( )( )
x
x
x
x
x
x
x
→
⋅
⋅
⋅






=
=
0
2
2
2
2
2
2
8
1
1 1 1 8
8 8
Answers
201–300