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# FP3 Limits Q

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1. Q:

My working thus far:

then

then

and I am pretty much stuck from here.

Thnx
2. (Original post by member910132)
Q:

My working thus far:

then

then

and I am pretty much stuck from here.

Thnx
I don't think that particular expansion helps a lot - in fact, it's not even valid as you're considering large x whilst it's only valid for . I would start again and multiply the limit by .
3. Trick: let x= 1/y.
4. (Original post by Farhan.Hanif93)
I don't think that particular expansion helps a lot - in fact, it's not even valid as you're considering large x whilst it's only valid for . I would start again and multiply the limit by .
But then I end up with

and I can't expand the bottom again because
and the expansion won't be valid for that.
Trick: let x= 1/y.
If I go ahead with this then does the limit change to
?

As then I can validly expand ?
6. (Original post by member910132)
If I go ahead with this then does the limit change to
?
Yes that becomes the limit.

Then you get:

(1/y^2 + 3/y)^1/2 -> (1/y^2(1+ 3y))^1/2

->((1/y)^2(1+ 3y))^1/2 -1/y it'll drop out fairly easily now.
7. (Original post by member910132)
But then I end up with

and I can't expand the bottom again because
and the expansion won't be valid for that.
Note that and bring that new factor of into the square root in the denominator. Things do simplify to a position to take the limit.

If you wanted to use binomial expansion from the start, you could have noted , which is valid for and hence will do the trick.
8. (Original post by Farhan.Hanif93)
Note that and bring that new factor of into the square root in the denominator. Things do simplify to a position to take the limit.

If you wanted to use binomial expansion from the start, you could have noted , which is valid for and hence will do the trick.
If then

so how did you get

?
9. (Original post by member910132)
If then

so how did you get

?
When taking the reciprocal, you have to reverse the inequality signs.
Yes that becomes the limit.

Then you get:

(1/y^2 + 3/y)^1/2 -> (1/y^2(1+ 3y))^1/2

->((1/y)^2(1+ 3y))^1/2 -1/y it'll drop out fairly easily now.
I just have a few concerns,

This is my first time doing FP3 and so I don't have the mathematical intuition you guys have, so how on earth was I supposed to think of

even after that, we get

which is fine, but I could have turned that into either of

and you will only get the correct answer of if you chose the second one.

So my point is how on earth was I supposed to think of all of that ? Is it likely the question would have at least said let ? F1, why dod you think of that substitution ? What made you think of it ?

Secondly then, should I make a habit of remembering that if then the expansions only valid for can't be used ? In fact, F1 is it this that made you use the substitution ? So you could justify the binomial series ?

(Original post by Farhan.Hanif93)
Note that and bring that new factor of into the square root in the denominator. Things do simplify to a position to take the limit.

If you wanted to use binomial expansion from the start, you could have noted , which is valid for and hence will do the trick.
Could you go into the "Note that and bring that new factor of into the square root in the denominator." in more detail, I don't really understand ?
When taking the reciprocal, you have to reverse the inequality signs.
So do the sings change when:

1. taking the reciprocals

2. multiplying/dividing by -ve number ?

Any others I should be aware of ?

Edit:

but how can something be smaller than -1 and bigger than 1 ?
12. (Original post by member910132)
If then

so how did you get

?

(Original post by member910132)
Could you go into the "Note that and bring that new factor of into the square root in the denominator." in more detail, I don't really understand ?
So we have .

Can you take the limit now?
13. (Original post by Farhan.Hanif93)

Can you do that without the modulus sign ? Showing the both sides though.

And how can be smaller than -1 and grater than 1?
14. (Original post by Farhan.Hanif93)

If you wanted to use binomial expansion from the start, you could have noted , which is valid for and hence will do the trick.
Leaving aside this confusion of mine about the range of validity,

after expanding it all I get to:

and that doesn't give me the answer of 3/2 as x -> infty.
15. (Original post by Farhan.Hanif93)

So we have .

Can you take the limit now?
Yea I got it form here, that is clever thinking though !!
16. (Original post by member910132)
Leaving aside this confusion of mine about the range of validity,

after expanding it all I get to:

and that doesn't give me the answer of 3/2 as x -> infty.
That's not quite correct. It looks like you've forgotten about the factor of x on the outside of the binomial expansion. Once you multiply that in, the result follows.
17. Right, and to save me from making a brand new thread, can someone address the issue of flipping the inequality sign when taking reciprocals ?

but how can 1/x be bigger then 1 and smaller than -1 ?

Or is it like this

?

Edit: Would and cannot it not be written as one inequality ?

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