Lim as n goes to inf of (sqrt(n)*sin((e^n)*pi))/(n+1)

lim as n goes to inf of (sqrt(n)*sin((e^n)*pi))/(n+1)

I tried looking up how to do it on Wolfram alpha and it has a very minimal amount of explanation and I really wanna know how to do these problems well. Can you explain the steps to this problem so I know how to do it? I guess doing a substitution helps ---> like let x=1/n. And on wolfram it goes to taking all of the limits directly. Is there more work to that or is that sufficient? Is the answer of 0 found simply in 2 lines?

Thank you so much for all of your help. I really appreciate the explanation. :)

Yes, it is easy to see the answer is zero. The sin(e^n*pi) is just a distraction, because the sin() of anything is always bounded between -1 and 1. So let f(n) be your function. We have for all n>0:

-sqrt(n)/(n+1) <= f(n) <= sqrt(n)/(n+1)

Then taking the limit of both sides sandwiches the f(n) function by 0, so the limit of f(n) as n goes to infinity must also be zero.