a)
#lim_(x->oo) e^(-2x)sqrtx#
This reads as #0*oo#. You can rewrite this to read as #0/0# or #oo/oo#, respectively:
#lim_(x->oo)e^(-2x)/(1/(sqrtx))#
or
#lim_(x->oo)sqrtx/(e^(2x))#
Now you can try to use L'Hopital's Rule to differentiate the numerator and denominator until it looks nice. Let's try the second one because #d/(dx)[1/(sqrtx)]# won't get you anywhere. (Getting these functions to shift to the bottom or top at once would be great.)
#= lim_(x->oo) 1/(2sqrtx*2e^(2x)) = lim_(x->oo) 1/(4sqrtxe^(2x))#
There we go. We're done.
#lim_(x->oo) e^(-2x)sqrtx = lim_(x->oo) 1/(4sqrtxe^(2x)) = 1/(ooe^(oo)) = 0#
b)
#lim_(x->oo) sin(6x)/(sin(7x))#
This one simply has no answer. #sinu# constantly oscillates, and if you keep moving towards #oo#, you'll never reach the end of the oscillation. Thus, the "answer" is simply #x in RR# (x is in the set of real numbers), or #(-oo,oo)#.
