Rewrite:
lim_(x->oo)root(4)xe^(-2x)=lim_(x->oo)x^(1/4)/e^(2x) (Because e^-x=1/e^x, root(a)x=x^(1/a))
Let's evaluate lim_(x->oo)x^(1/4)/e^(2x) straight away and see what happens:
lim_(x->oo)x^(1/4)/e^(2x)=oo^(1/4)/(e^(2*oo))=oo/oo
We have an indeterminate form; thus, we must apply l'Hopital's Rule, which tells us that
lim_(x->a)f(x)/g(x)=lim_(x->a)(f'(x))/(g'(x))
whenever lim_(x->a)f(x)/g(x) yields an indeterminate form.
For lim_(x->oo)x^(1/4)/e^(2x), we see f(x)=x^(1/4), g(x)=e^(2x). Differentiating, we get
f'(x)=1/4x^(-3/4), g'(x)=2e^(2x)
We can now rewrite our limit with l'Hopital's Rule applied:
lim_(x->oo)x^(1/4)/e^(2x)=lim_(x->oo)(1/4x^(-3/4))/(2e^(2x))=lim_(x->oo)(1/4)/(2x^(3/4)e^(2x)) (Because x^-a=1/x^a, we move our negative exponent to the denominator)
Let's evaluate:
lim_(x->oo)(1/4)/(2x^(3/4)e^(2x))=(1/4)/(2(oo)^(3/4)e^(2*oo))=(1/4)/oo=0
(Because any constant value divided by an extremely large value yields 0. a/oo=0.)
