Finding the limit of a definite integral

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Evaluate:
$$
operatorname*Lim_x to + infty
fracmathrmdmathrmdx
int_2 sin frac1x^3 sqrtx
frac3 t^4 + 1(t-3)(t^2 + 3)
,mathrmdt
$$




I have tried applying the Newton-Leb rule to the integral which becomes a really big equation and the there $mathrmd/mathrmdx$ of that. I was thinking of applying L’Hospital rule but I’m not sure if I’ll get a right answer with such a big equation. Is there any other way to solve it?




differential-equations limits definite-integrals




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  • The main tool you have to apply is the fundamental theorem of calculus: $fracddxint_a^x f(t),dt=f(x)$.
    – egreg
    Aug 22 at 9:21






  • 2




    But $t$ crosses $t=3$ so is it even defined ?
    – Empy2
    Aug 22 at 9:29










  • @egreg so theres no use of the limits?
    – user568101
    Aug 22 at 9:36










  • @user568101 Of course they have to be used.
    – egreg
    Aug 22 at 9:46






  • 1




    I think @Empy2 is correct. The function is undefined.
    – KittyL
    Aug 22 at 10:26














up vote
0
down vote

favorite













Evaluate:
$$
operatorname*Lim_x to + infty
fracmathrmdmathrmdx
int_2 sin frac1x^3 sqrtx
frac3 t^4 + 1(t-3)(t^2 + 3)
,mathrmdt
$$




I have tried applying the Newton-Leb rule to the integral which becomes a really big equation and the there $mathrmd/mathrmdx$ of that. I was thinking of applying L’Hospital rule but I’m not sure if I’ll get a right answer with such a big equation. Is there any other way to solve it?




differential-equations limits definite-integrals




share|cite|improve this question






















  • The main tool you have to apply is the fundamental theorem of calculus: $fracddxint_a^x f(t),dt=f(x)$.
    – egreg
    Aug 22 at 9:21






  • 2




    But $t$ crosses $t=3$ so is it even defined ?
    – Empy2
    Aug 22 at 9:29










  • @egreg so theres no use of the limits?
    – user568101
    Aug 22 at 9:36










  • @user568101 Of course they have to be used.
    – egreg
    Aug 22 at 9:46






  • 1




    I think @Empy2 is correct. The function is undefined.
    – KittyL
    Aug 22 at 10:26












up vote
0
down vote

favorite









up vote
0
down vote

favorite












Evaluate:
$$
operatorname*Lim_x to + infty
fracmathrmdmathrmdx
int_2 sin frac1x^3 sqrtx
frac3 t^4 + 1(t-3)(t^2 + 3)
,mathrmdt
$$




I have tried applying the Newton-Leb rule to the integral which becomes a really big equation and the there $mathrmd/mathrmdx$ of that. I was thinking of applying L’Hospital rule but I’m not sure if I’ll get a right answer with such a big equation. Is there any other way to solve it?




differential-equations limits definite-integrals




share|cite|improve this question















Evaluate:
$$
operatorname*Lim_x to + infty
fracmathrmdmathrmdx
int_2 sin frac1x^3 sqrtx
frac3 t^4 + 1(t-3)(t^2 + 3)
,mathrmdt
$$




I have tried applying the Newton-Leb rule to the integral which becomes a really big equation and the there $mathrmd/mathrmdx$ of that. I was thinking of applying L’Hospital rule but I’m not sure if I’ll get a right answer with such a big equation. Is there any other way to solve it?





differential-equations limits definite-integrals





share|cite|improve this question













share|cite|improve this question




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edited Aug 23 at 10:45









Jendrik Stelzner

7,57221037




7,57221037










asked Aug 22 at 9:06









user568101

185




185











  • The main tool you have to apply is the fundamental theorem of calculus: $fracddxint_a^x f(t),dt=f(x)$.
    – egreg
    Aug 22 at 9:21






  • 2




    But $t$ crosses $t=3$ so is it even defined ?
    – Empy2
    Aug 22 at 9:29










  • @egreg so theres no use of the limits?
    – user568101
    Aug 22 at 9:36










  • @user568101 Of course they have to be used.
    – egreg
    Aug 22 at 9:46






  • 1




    I think @Empy2 is correct. The function is undefined.
    – KittyL
    Aug 22 at 10:26
















  • The main tool you have to apply is the fundamental theorem of calculus: $fracddxint_a^x f(t),dt=f(x)$.
    – egreg
    Aug 22 at 9:21






  • 2




    But $t$ crosses $t=3$ so is it even defined ?
    – Empy2
    Aug 22 at 9:29










  • @egreg so theres no use of the limits?
    – user568101
    Aug 22 at 9:36










  • @user568101 Of course they have to be used.
    – egreg
    Aug 22 at 9:46






  • 1




    I think @Empy2 is correct. The function is undefined.
    – KittyL
    Aug 22 at 10:26















The main tool you have to apply is the fundamental theorem of calculus: $fracddxint_a^x f(t),dt=f(x)$.
– egreg
Aug 22 at 9:21




The main tool you have to apply is the fundamental theorem of calculus: $fracddxint_a^x f(t),dt=f(x)$.
– egreg
Aug 22 at 9:21




2




2




But $t$ crosses $t=3$ so is it even defined ?
– Empy2
Aug 22 at 9:29




But $t$ crosses $t=3$ so is it even defined ?
– Empy2
Aug 22 at 9:29












@egreg so theres no use of the limits?
– user568101
Aug 22 at 9:36




@egreg so theres no use of the limits?
– user568101
Aug 22 at 9:36












@user568101 Of course they have to be used.
– egreg
Aug 22 at 9:46




@user568101 Of course they have to be used.
– egreg
Aug 22 at 9:46




1




1




I think @Empy2 is correct. The function is undefined.
– KittyL
Aug 22 at 10:26




I think @Empy2 is correct. The function is undefined.
– KittyL
Aug 22 at 10:26










1 Answer
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As you know that according to Leibnitz theorem for integrls
$$fracddx,int_a(x)^b(x) f(t),dt=f(b(x)), b'(x)-f(a(x)) ,a'(x)$$




$$fracdIdx= lim_x to inftyleft(frac3^5 x^2+19(sqrtx-1)(3x^2+1)cdotfrac32sqrtx-frac3cdot2^4sin^4(1/x)+1(2sin(1/x)-3)(4sin^2(1/x)+3)cdot2 cos(1/x)(-frac1x^2)right)$$







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    up vote
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    down vote













    As you know that according to Leibnitz theorem for integrls
    $$fracddx,int_a(x)^b(x) f(t),dt=f(b(x)), b'(x)-f(a(x)) ,a'(x)$$




    $$fracdIdx= lim_x to inftyleft(frac3^5 x^2+19(sqrtx-1)(3x^2+1)cdotfrac32sqrtx-frac3cdot2^4sin^4(1/x)+1(2sin(1/x)-3)(4sin^2(1/x)+3)cdot2 cos(1/x)(-frac1x^2)right)$$







    share|cite|improve this answer
























      up vote
      0
      down vote













      As you know that according to Leibnitz theorem for integrls
      $$fracddx,int_a(x)^b(x) f(t),dt=f(b(x)), b'(x)-f(a(x)) ,a'(x)$$




      $$fracdIdx= lim_x to inftyleft(frac3^5 x^2+19(sqrtx-1)(3x^2+1)cdotfrac32sqrtx-frac3cdot2^4sin^4(1/x)+1(2sin(1/x)-3)(4sin^2(1/x)+3)cdot2 cos(1/x)(-frac1x^2)right)$$







      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        As you know that according to Leibnitz theorem for integrls
        $$fracddx,int_a(x)^b(x) f(t),dt=f(b(x)), b'(x)-f(a(x)) ,a'(x)$$




        $$fracdIdx= lim_x to inftyleft(frac3^5 x^2+19(sqrtx-1)(3x^2+1)cdotfrac32sqrtx-frac3cdot2^4sin^4(1/x)+1(2sin(1/x)-3)(4sin^2(1/x)+3)cdot2 cos(1/x)(-frac1x^2)right)$$







        share|cite|improve this answer












        As you know that according to Leibnitz theorem for integrls
        $$fracddx,int_a(x)^b(x) f(t),dt=f(b(x)), b'(x)-f(a(x)) ,a'(x)$$




        $$fracdIdx= lim_x to inftyleft(frac3^5 x^2+19(sqrtx-1)(3x^2+1)cdotfrac32sqrtx-frac3cdot2^4sin^4(1/x)+1(2sin(1/x)-3)(4sin^2(1/x)+3)cdot2 cos(1/x)(-frac1x^2)right)$$








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 22 at 11:29









        Deepesh Meena

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