Part II: The Answers
190
Answers
101–200
181.
108
Note that substituting the limiting value, 3, into the function x
x
4
81
3
−
−
gives you the indeterminate form 0
0
.
To find the limit, first factor the numerator and simplify:
lim
lim
lim
(
)(
)
x
x
x
x
x
x
x
x
x
x
x
→
→
→
−
−
=
−
(
) +
(
)
−
=
−
+
+
(
3
4
3
2
2
3
2
81
3
9
9
3
3
3
9 ) )
−
=
+
+
(
)
→
x
x
x
x
3
3
9
3
2
lim(
)
Then substitute 3 for x:
lim(
)
(
)
( )( )
x
x
x
→
+
+
(
)
= +
+
(
)
=
=
3
2
2
3
9
3 3 3 9
6 18
108
182.
∞
Note that substituting in the limiting value gives you an indeterminate form.
Because x is approaching 0 from the right, you have x > 0 so that x x
= . Therefore, the
limit becomes
lim
lim
lim
x
x
x
x
x
x
x
x
x
→
→
→
+
+
+
+


 


  =
+






=
+



0
2
0
2
0
2
2
2
2
2
2 2  


As x → 0
+
, you have (2 + 2x) → 2 and x
2 → 0
+
so that 2 2
2
0
2
+





 →
→∞
+
x
x
. The limit
is positive infinity because dividing 2 by a very small positive number close to zero
gives you a very large positive number.
183.
∞
Note that substituting in the limiting value gives you the indeterminate form ∞ – ∞.
Because x is approaching 0 from the left, you have x < 0 so that x
x
= − . Therefore, the
limit becomes