evaluate limit

Since 1<=sin^2(2pi/x)+1<=2 for all x and since lim x->0 sqrt(x) = 0 you will have that limx->0 sqrt(x)*(sin^2(2pi/x)) = 0

Note: Have in mind that limx->+0 2pi/x = infinity, so sin^2(infinity) will be between 0 and 2, but you then have a +1 so it will be between 1 and 2. However since the lim from sqrt will be 0 you can "say" that it is 0*sin^2(2pi/x)+1.

Hope it helps!

Hi cristobalsele, thank you for your  good explanation, I don't quite understand why 1<=  and x->0. I don't understand which rule I've to apply and how to take the first step etc.

So normally a function for example like f(x)=x or f(x)=x^2 and evaluate it's limit it tends to go into a specific value, in bot examples it would be lim x->inf = inf. However when you evaluate functions like sin(x) or cos(x), you can see by graphing and following them to infinity both functions will be between 1 and -1. In such cases you have to use the so called "Sandwich Theorem". 

I recommend you to look for a youtube video that explains exactly how to use this theorem and after that you should have a better approach to such example.