# Real analysis limits

1. Nov 20, 2008

### squaremeplz

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!!

Last edited: Nov 21, 2008
2. Nov 21, 2008

### sutupidmath

oK, lets use proof by contradiction, that is say, that even though $$f_1<f_2$$ for all values on (a,b) we will have $$L_2<L_1$$

Now,

$$\lim_{x \rightarrow a+}[f_2(x)-f_1(x)]=L_2-L_1$$

So $$\forall \epsilon>0$$ also for $$\epsilon =L_1-L_2, \exists \delta$$ such that whenever

$$a<x<a+\delta => |f_2(x)-f_1(x)-(L_2-L_1)|<L_1-L_2$$ but since we know that

$$a\leq|a|$$ we get

$$f_2(x)-f_1(x)-(L_2-L_1)\leq |f_2(x)-f_1(x)-(L_2-L_1)|<L_1-L_2=>f_2(x)-f_1(x)<0=>f_2(x)<f_1(x)$$ which clearly is not possible, since it violates our initial hypothesis.

so $$L_1<L_2$$

3. Nov 21, 2008

thank you!