Answers and Explanations 203
213.
–∞
As x approaches 1 from the left, you have (x – 1) → 0
–
so that
3
1
3
0
x −
→
→−∞
−
Note that the limit is negative infinity because dividing 3 by a small negative number
close to zero gives you a negative number whose absolute value is large.
214.
–∞
Begin by writing the limit as lim (tan ) lim sin
cos
x
x
x
x
x
→
→
+
+
=
π
π
2
2
. Then consider what
happens in the numerator and the denominator as x approaches π
2
from the right.
As x →
+
π
2
, you have sin x → 1 and cos x → 0
–
. Therefore, as x →
+
π
2
, it follows that
sin
cos
x
x
→
→−∞
−
1
0
215.
∞
Consider what happens in the numerator and denominator as x approaches π from the
left. As x → π
 –
, you have x
2 → π
2
and sin x → 0
+
. Therefore, as x → π
 –
, it follows that
x
x
2
2
0
sin
→
→∞
+
π
216.
–∞
Consider what happens in the numerator and denominator as x approaches 5 from the
left. As x → 5
–
, you have (x + 3) → 8 and (x – 5) → 0
–
. Therefore, as x → 5
–
, it follows that
x
x
+
−
→
→−∞
−
3
5
8
0
Answers
201–300