Evaluate the limit? #lim_(x rarr 0) x^2(cos(1/x)-1) #
2 Answers
# lim_(x rarr 0) x^2(cos(1/x)-1) = 0#
Explanation:
We have:
# L = lim_(x rarr 0) x^2(cos(1/x)-1) #
Let us perform a substitution:
Let
#u=1/x => x =1/u#
Then as
Substituting into the limit, we get:
# L = lim_(u rarr oo) (1/u)^2(cos(u)-1) #
# \ \ = lim_(u rarr oo) (cosu-1)/u^2 #
# \ \ = lim_(u rarr oo) (cosu)/u^2-1/u^2 #
# \ \ = lim_(u rarr oo) (cosu)/u^2- lim_(u rarr oo)1/u^2 #
Consider the second limit:
# lim_(u rarr oo)1/u^2 #
This is a trivial limit as
# lim_(u rarr oo)1/u^2 = 0 #
Consider now, the first limit:
# lim_(u rarr oo) (cosu)/u^2 #
Now, we have
# lim_(u rarr oo)(-1/u^2) le lim_(u rarr oo)(cosu /u^2) le lim_(u rarr oo)(1/u^2) #
And using the above result we just established that:
# 0 le lim_(u rarr oo)(cosu /u^2) le 0 #
So then by the sandwich, or squeeze, theorem we have:
# lim_(u rarr oo)(cosu /u^2) = 0 #
Thus:
# L = 0 #
I would use the squeeze theorem.
Explanation:
So , subtracting
Note that
Since
the squeeze theorem tells us that