Part II: The Answers
194
Answers
101–200
189.
–1
Note that substituting in the limiting value gives you the indeterminate form 0
0
.
Because x is approaching 5 from the left, you have x < 5 so that x
x
− = − −
5
5
(
).
Therefore, the limit becomes
lim
lim (
)
lim ( )
x
x
x
x
x
x
x
→
→
→
−
−
−
−
−
=
−
− −
=
−
= −
5
5
5
5
5
5
5
1
1
190.
− 1
25
Note that substituting in the limiting value, –5, into the function
1 5
1
5
+
+
x
x
gives you the
indeterminate form 0
0
.
Begin by writing the two fractions in the numerator as a single fraction by getting
common denominators. Then simplify:
lim
lim
( )
( )
( )
( )
lim
x
x
x
x
x
x
x
x
x
x
x
x
→−
→−
→−
+
+
=
+
+
=
+
+
5
5
5
1
5
1
5
1
5
1 5
5
5
5
5
5 5
1
5
5
1
5
1
5
5
5
=
+
( ) +
( )
=
→−
→−
lim
lim
x
x
x
x
x
x