Question

Question #5f39e

Answer 1

`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`