Evaluate the limit? #lim_(x rarr 0) x^2(cos(1/x)-1) #

2 Answers
Aug 6, 2017

# 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 #x rarr 0 => u rarr oo #

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 #u^2# increases without bound, thus #1/u^2 rarr 0# as #u rarr oo#, thus:

# lim_(u rarr oo)1/u^2 = 0 #

Consider now, the first limit:

# lim_(u rarr oo) (cosu)/u^2 #

Now, we have #-1 le cosu le 1 => -1/u^2 le cosu /u^2 le 1/u^2 #, thus:

# 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 #

Aug 6, 2017

I would use the squeeze theorem.

Explanation:

#-1<=cos(1/x)<=1# #" "# for all #x != 0#

So , subtracting #1# from each part,

#-2 <= cos(1/x)-1 <= 0# #" "# for all #x != 0#

Note that #x^2 >0# #" "# for all #x != 0#, so we can multiply through without changing the inequalities.

#-2x^2 <= x^2(cos(1/x)-1 ) <= 0# #" "# for all #x != 0#

Since #lim_(xrarr0)(-2x^2) = 0 = lim_(xrarr0)0#

the squeeze theorem tells us that

#Lim_(xrarr0)x^2(cos(1/x)-1 ) = 0#