Find lim x-> infinite. f(x) = (3x/(e^2x + 7x^2))^(1/x)

We are asked to find lim_(x->infty) ((3x)/(e^(2x)+7x^2))^(1/x) :
First, use the fact that the limit of a product is the product of limits to get:
=lim_(x->infty)3^(1/x) * lim_(x->infty)((x)/(e^(2x)+7x^2))^(1/x)
Note that lim_(x->infty)3^(1/x)=3^(lim_(x->infty)1/x)=3^0=1
Now for the remaining factor write as:
=lim_(x->infty)e^(ln((x)/(e^(2x)+7x^2))^(1/x)
Use a property of logarithms (ln is the natural log function) to get:
=lim_(x->infty)e^((ln((x)/(e^(2x)+7x^2))/x)
=e^(lim_(x->infty)((ln((x)/(e^(2x)+7x^2))/x)
The limit is indeterminate, so use L'hopital's rule:
d/(dx) (ln((x)/(e^(2x)+7x^2))/x)=-(e^(-2x)+2e^(2x)+7x^2)/(x(e^(2x)+7x^2))
so we have
=e^(lim_(x->infty) -(e^(-2x)+2e^(2x)+7x^2)/(x(e^(2x)+7x^2))
=e^(-lim_(x->infty)(e^(-2x)+2e^(2x)+7x^2)/(xe^(2x)+7x^3)
Divide the argument through by xe^(2x) to get:

e^(-2/1)=1/e^2