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.
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