260 CHAPTER 5 Series and Differential Equations
Other indeterminate forms can be treated as in the following exam-
ples.
Example 2. Compute limx→∞
(
1 −4
x)x. Here, if we set Lequal to this
limit (if it exists!), then we have, by continuity of the logarithm, that
lnL = ln limx→∞(
1 −4
x)x= limx→∞ln(
1 −4
x)x= limx→∞xln(
1 −4
x)= limx→∞ln(
1 −x^4)1 /x
l’H= lim
x→∞4 /
(
x^2(
1 −x^4))− 1 /x^2
= limx→∞( −^4
1 −^4 x)=− 4
This says that lnL=−4 which implies thatL=e−^4.
Example 3. This time, try lim
θ→(π/2)−
(cosθ)cosθ. The same trick applied
above works here as well. SettingLto be this limit, we have
lnL = ln lim
θ→(π/2)−(cosθ)cosθ= lim
θ→(π/2)−
ln(cosθ)cosθ
= lim
θ→(π/2)−cosθln cosθ= lim
θ→(π/2)−ln cosθ
1 /cosθ
l’H= lim
θ→(π/2)−tanθ
secθtanθ= 0.
It follows that lim
θ→(π/2)−
(cosθ)cosθ= 1.