Real Analysis Limits Homework: Show L1<=L2 & Find lim(x->0) f(x)

[squaremeplz]

Homework Statement

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.

Homework Equations

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!

 

[sutupidmath]

Homework Statement

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

oK, let's use proof by contradiction, that is say, that even though f_1&lt;f_2 for all values on (a,b) we will have L_2&lt;L_1

Now,

<br /> \lim_{x \rightarrow a+}[f_2(x)-f_1(x)]=L_2-L_1

So \forall \epsilon&gt;0 also for \epsilon =L_1-L_2, \exists \delta such that whenever

a&lt;x&lt;a+\delta =&gt; |f_2(x)-f_1(x)-(L_2-L_1)|&lt;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)|&lt;L_1-L_2=&gt;f_2(x)-f_1(x)&lt;0=&gt;f_2(x)&lt;f_1(x) which clearly is not possible, since it violates our initial hypothesis.

so L_1&lt;L_2

 

[squaremeplz]

thank you!