1. The problem statement, all variables and given/known data 1. Suppose that L1 = lim(x->a+) f1(x) and L2 = lim(x->a+) f2(x) Also, suppose that f1(x)<=f2(x) for all x in (a,b). Show that L1<=L2 2. Suppose f(x) = (sqrt(1+3*x^2) - 1)/(x^2) show that the lim (x->0) f(x) exits and give its value. 2. Relevant equations 3. The attempt at a solution 1) I find this problem a tad cumbersome for I have no clue as to where to start. I drew a graph which led me to the following conclusion for any sequence x_n in (a,b) 1 [lim(n) f1(x_n) = f1(x_o)] <= [lim(n) f2(x_n) = f2(x_o)] 2 and maybe |L2 - L1| = |f2(x) - f1(x)| but 1 shows that when x_n gets arbitrarily close to a, f1(a)<=f2(a) 2) I multiplied the top of f(x) by (sqrt(1+3*x^2) + 1). that is (sqrt(1+3*x^2) - 1)/(x^2) * (sqrt(1+3*x^2) + 1) = 3. hence the lim(x->0) f(x) = 3 thanks!!