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# a interesting and difficult question about limits and integration watch

1. lim x goes to 0()int from o to x (x^2sinx dx/x^2)
2. Let me post a picture
3. (Original post by ElliotWalton)
...
Not that interesting. Just apply L'Hopitals twice?

Alternatively can be done by just expanding using MacLaurin's then integrating the polynomial before dividing by and doing the limit
4. (Original post by RDKGames)
Not that interesting. Just apply L'Hopitals twice?

Alternatively can be done by just expanding using MacLaurin's then integrating the polynomial before dividing by and doing the limit
5. (Original post by ElliotWalton)
Do you understand how L'Hopital's rule works?
6. (Original post by RDKGames)
Do you understand how L'Hopital's rule works?
absolutely i know that
7. Actually, the question isn't quite correct. You can't have x both in the integration limit and as an integration variable.
8. (Original post by ElliotWalton)
absolutely i know that
What Zacken said is correct and I didn't pick up on it - perhaps the question should read ?
9. (Original post by RDKGames)
What Zacken said is correct and I didn't pick up on it - perhaps the question should read ?
10. (Original post by ElliotWalton)
Then in that case you should know that applying L'Hopitals would mean you need , but what does the Fundamental Theorem of Calculus tells us this is?
11. (Original post by RDKGames)
Then in that case you should know that applying L'Hopitals would mean you need , but what does the Fundamental Theorem of Calculus tells us this is?
Frankly speaking,I have no idea.
12. (Original post by RDKGames)
Not that interesting. Just apply L'Hopitals twice?
Not sure there's even any need to apply L'Hopital - just bound the integrand by x^2.
13. (Original post by ElliotWalton)
Frankly speaking,I have no idea.
Quick wikipedia look at the theorem says that for , we have so you can just apply this.
14. (Original post by DFranklin)
Not sure there's even any need to apply L'Hopital - just bound the integrand by x^2.
Ah didn't notice that - not used to bounding these, but there's the third method for OP then.
15. (Original post by RDKGames)
Quick wikipedia look at the theorem says that for , we have so you can just apply this.
Thanks for your help.Let me have a try.BTW,would you mind showing a complete working of this question?
16. You don't need to calculate the whole integral to get the answer, you can just use the inequality sin t <= 1 to get that the absolute value of the integral is less than x/3 hence tends to 0 as x tends to 0.
17. (Original post by Jeremy01)
~snip~
Please see the Posting Guidelines and in particular what they have to say about posting full solutions.

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