what is the limit as x approaches infinity of (x^3)*(sin(2pi/x))? and how to do it please

Question

I keep trying to solve it by using l'hopitals rule but I can't seem to get the answer right if anyone could help me out not only with the answer but a step by step explaination of how to work this problem thanks in advance

Answer 1

This shouls answer your question:

http://www.wolframalpha.com/input/?i=lim%28x-%3Einfinity%29+%28x%5E3%29*sin%282pi+%2F+x%29

 

Click on the "Show Steps" button on the right hand side.

Answer 2

You need to use squeeze theorem to answer this questions.  Breaking the function into parts, the lim as x → 0 of 2pi/x is infinity, so we have sine function going to infinity which is alternating between -1 and +1 or -1 < sin(2pi/x) < 1  now the lim x → 0 of x^3 is 0 so when you multiply 0 times the +/- 1 you get 0.

Therefore lim x→0 x^3 sin(2pi/x) = 0