As typed, the question asks for #lim_(xrarr1)(x-1)^2 cos(1/x-1)#.
Interpretation 1
If you really want #lim_(xrarr1)(x-1)^2 cos(1/x-1)#, use substitution to get
#(1-1)^2 cos(1/1-1) = 0cos(0) = 0*1 = 0#
Interpretation 2
For #lim_(xrarr1) (x-1)^2 cos(1/(x-1))# argue something like this:
We know from our study of trigonometry, that
#-1 <= cos(1/(x-1)) <= 1# for all #x != 1#.
Since #(x-1)^2# is positive for all #x != 1#, we can multiply the inequality without changing the direction of the inequality.
#-(x-1)^2 <= (x-1)^2 cos(1/(x-1)) <= (x-1)^2# for #x != 1#
Note that
#lim_(xrarr1)(-(x-1)^2) = 0# and #lim_(xrarr1)(x-1)^2 = 0#.
Therefore, by the squeeze theorem,
#lim_(xrarr1) (x-1)^2 cos(1/(x-1))#
