Let #L = lim_(x->oo) e^sqrtx/sqrt(e^x+1)#.
Since we have a square root, it'd be easier to calculate the limit of the function squared. In other words,
#L^2 = lim_(x->oo) (e^sqrtx/sqrt(e^x+1))^2 = lim_(x->oo) e^(2sqrtx)/(e^x+1)#
Let's rewrite this as the following:
#L^2 = lim_(x->oo) e^(2sqrtx)/e^x * color(red)(e^x/(e^x+1)#
If you do the multiplication, it's the same as before.
#L^2 = lim_(x->oo) e^(2sqrtx-x) * color(red)(lim_(x->oo)e^x/(e^x+1)#
The second limit is, intuitively, equal to #1#, and you can prove it by using #"L'Hôpital's rule"#.
#lim_(x->oo) e^x/(e^x+1) = (d/dx e^x)/(d/dx e^x+1) = e^x/e^x = 1#.
Therefore,
#L^2 = lim_(x->oo) e^(2sqrtx-x) = e^(lim_(x->oo)2sqrtx-x)#
#L^2 = e^(lim_(x->oo) sqrtx(2-sqrtx)) = e^(oo*(-oo)) = e^(-oo) =0#
#:. L =0#.
