Limits and Continuity 51Example 2
Find the limit: limx→π 3 xsinx.
Using the product rule, limx→π 3 xsinx=
(
xlim→π^3 x)(
xlim→πxlim→πsinx)
=(3π)(sinπ)=(3π)(0)=Example 3
Find the limit: limt→ 2
t^2 − 3 t+ 2
t− 2
.
Factoring and simplifying: limt→ 2
t^2 − 3 t+ 2
t− 2
=limt→ 2(t−1)(t−2)
(t−2)=limt→ 2 (t−1)=(2−1)= 1.(Note that had you substitutedt=2 directly in the original expression, you would have
obtained a zero in both the numerator and denominator.)
Example 4
Find the limit: limx→b
x^5 −b^5
x^10 −b^10
.
Factoring and simplifying: limx→b
x^5 −b^5
x^10 −b^10
=limx→b
x^5 −b^5
(x^5 −b^5 )(x^5 +b^5 )
=xlim→b1
x^5 +b^5=
1
b^5 +b^5=
1
2 b^5.
Example 5
Find the limit: limt→ 0
√
t+ 2 −√
2
t.
Multiplying both the numerator and the denominator by the conjugate of the numerator,
(√
t+ 2 +
√
2)
, yields limt→ 0√
t+ 2 −√
2
t(√
t+ 2 +√
2
√
t+ 2 +√
2)=limt→ 0
t+ 2 − 2
t
(√
t+ 2 +√
2)=limt→ 0
t
t
(√
t+ 2 +√
2)=limt→ 01
(√
t+ 2 +√
2)=1
√
0 + 2 +√
2=
1
2
√
2=
1
2
√
2(√
2
√
2)
=√
2
4.
(Note that substituting 0 directly into the original expression would have produced a 0 in
both the numerator and denominator.)