Question #8cc23

1 Answer
Jim H
Nov 30, 2016

Please see below.

Explanation:

#-1 <= sin(pi/x) <= 1# for all #x != 0#.

#sqrt(x^3+x^2) > 0# so we can multiply without changing the inequalities.

#-sqrt(x^3+x^2) <= sqrt(x^3+x^2)sin(pi/x) <= sqrt(x^3+x^2)# #" "# for all #x != 0#.

Observe that #lim_(xrarr0)-sqrt(x^3+x^2) = 0# and #lim_(xrarr0)sqrt(x^3+x^2) = 0#.

Therefore, by the squeeze theorem, #lim_(xrarr0)sqrt(x^3+x^2)sin(pi/x) = 0#

Related questions
See all questions in Introduction to Limits
Impact of this question
2948 views around the world
You can reuse this answer
Creative Commons License