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Real analysis limits

  1. Nov 20, 2008 #1
    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. jcsd
  3. Nov 21, 2008 #2


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

    Now,

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

    So [tex]\forall \epsilon>0[/tex] also for [tex] \epsilon =L_1-L_2, \exists \delta[/tex] such that whenever

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

    [tex] a\leq|a|[/tex] we get

    [tex]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)[/tex] which clearly is not possible, since it violates our initial hypothesis.

    so [tex]L_1<L_2[/tex]
     
  4. Nov 21, 2008 #3
    thank you!
     
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