Question #5f39e

1 Answer
Jul 19, 2017

lim_"x->0"sqrtxe^sin(pi/x)=0

Explanation:

We know that -1<=sin(pi/x)<=1=>e^(-1)<=e^sin(pi/x)<=e=>

sqrt(x)/e<=sqrtxe^sin(pi/x)<=sqrtxe

lim_"x->0"sqrt(x)/e=0

lim_"x->0"sqrtxe=0

So we know from the squeeze theorem that

lim_"x->0"sqrtxe^sin(pi/x)=0