How do you use the Squeeze Theorem to find #lim(x-1)sin(pi/x-1) # as x approaches one?

2 Answers
Oct 11, 2015

#lim_(x rarr 1) (x-1)sin(pi/x - 1) = 0#

Explanation:

Firstly, we don't need the squeeze theorem because the function is continuous at values close to #x=1#, so the limit is #f(1)#, that being said, to use the squeeze theorem we must remember that

#-1 <= sin(theta) <= 1#

If we have #theta = pi/x - 1#, we have

#-1 <= sin(pi/x - 1) <= 1#

Multiplying both sides by #x-1# we have

#1 -x <= (x-1)sin(pi/x - 1) <= x - 1# for #x >0#

Since

#lim_(x rarr 1) 1 - x = lim_(x rarr 1) x - 1 = 0#

The squeeze theorem tells us that

#lim_(x rarr 1) (x-1)sin(pi/x - 1) = 0#

Oct 11, 2015

#lim_(xrarr1)(x-1)sin(pi/(x-1)) = 0#

Explanation:

#-1 <= sin(pi/(x-1)) <= 1# for all #x != 1#

Limit from the right is 0

For #x > 1#, we have #x-1 > 0# so we can multiply the inequality without changing the inequalities:

# -(x-1) <= (x-1)sin(pi/(x-1)) <= x-1# for all #x != 1#

#lim_(xrarr1^+) -(x-1) = 0# and #lim_(xrarr1^+) (x-1) = 0#

So #lim_(xrarr1^+) (x-1)sin(pi/(x-1)) = 0#

Limit from the left is 0

For #x < 1#, we have #x-1 < 0# so when we multiply the inequality we must change the inequalities:

# -(x-1) >= (x-1)sin(pi/(x-1)) >= x-1# for all #x != 1#

#lim_(xrarr1^-) -(x-1) = 0# and #lim_(xrarr1^-) (x-1) = 0#

So #lim_(xrarr1^-) (x-1)sin(pi/(x-1)) = 0#

Therefore,

#lim_(xrarr1) (x-1)sin(pi/(x-1)) = 0#