The Student Room Group

FP3 Limits Q

Q:

limx((x2+3x)12x) \displaystyle\lim_{x\to \infty} \left( (x^2 + 3x)^\frac{1}{2} - x \right)

My working thus far:

((x2+3x)12x)=(3x)12×(1+x3)12x \displaystyle \left( (x^2 + 3x)^\frac{1}{2} - x \right) = (3x)^\frac{1}{2} \times (1+ \frac {x}{3})^\frac{1}{2} - x

then

3x(1+x6x272+)x \displaystyle \sqrt {3x} \left ( 1 + \frac {x}{6} - \frac {x^2}{72} + \cdots \right) - x

then

(3x+3(x)3263(x)5272+)x \displaystyle \left ( \sqrt {3x} + \frac {\sqrt {3} (x)^\frac {3}{2}}{6} - \frac {\sqrt {3} (x)^\frac {5}{2}}{72} + \cdots \right) - x

and I am pretty much stuck from here.

Thnx :smile:
(edited 11 years ago)
Original post by member910132
Q:

limx((x2+3x)12x) \displaystyle lim_{x\to \infty} \left( (x^2 + 3x)^\frac{1}{2} - x \right)

My working thus far:

((x2+3x)12x)=(3x)12×(1+x3)12x \displaystyle \left( (x^2 + 3x)^\frac{1}{2} - x \right) = (3x)^\frac{1}{2} \times (1+ \frac {x}{3})^\frac{1}{2} - x

then

3x(1+x6x272+)x \displaystyle \sqrt {3x} \left ( 1 + \frac {x}{6} - \frac {x^2}{72} + \cdots \right) - x

then

(3x+3(x)3263(x)5272+)x \displaystyle \left ( \sqrt {3x} + \frac {\sqrt {3} (x)^\frac {3}{2}}{6} - \frac {\sqrt {3} (x)^\frac {5}{2}}{72} + \cdots \right) - x

and I am pretty much stuck from here.

Thnx :smile:

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 x<3|x|<3. I would start again and multiply the limit by x2+3x+xx2+3x+x\dfrac{\sqrt{x^2+3x} +x}{\sqrt{x^2+3x}+x}.
(edited 11 years ago)
Reply 2
Trick: let x= 1/y.
Reply 3
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 x<3|x|<3. I would start again and multiply the limit by x2+3x+xx2+3x+x\dfrac{\sqrt{x^2+3x} +x}{\sqrt{x^2+3x}+x}.


But then I end up with



3xx2+3x+x \displaystyle \frac {3x}{\sqrt {x^2 +3x} + x}

and I can't expand the bottom again because
x \displaystyle x \to \infty and the expansion won't be valid for that.
Reply 4
Original post by f1mad
Trick: let x= 1/y.


If I go ahead with this then does the limit change to
limy0 \displaystyle \lim_{y \to 0} ?

As then I can validly expand (1+13y)2 \displaystyle( 1+ \frac {1}{3y})^2 ?
Reply 5
Original post by member910132
If I go ahead with this then does the limit change to
limy0 \displaystyle \lim_{y \to 0} ?


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.
(edited 11 years ago)
Original post by member910132
But then I end up with



3xx2+3x+x \displaystyle \frac {3x}{\sqrt {x^2 +3x} + x}

and I can't expand the bottom again because
x \displaystyle x \to \infty and the expansion won't be valid for that.

Note that x11/xx\equiv \dfrac{1}{1/x} and bring that new factor of 1x\dfrac{1}{x} 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 (x2+3x)12x(1+3x)12\left(x^2+3x\right)^{\frac{1}{2}} \equiv x\left(1 + \dfrac{3}{x}\right)^{\frac{1}{2}}, which is valid for x>3|x|>3 and hence will do the trick.
(edited 11 years ago)
Reply 7
Original post by Farhan.Hanif93
Note that x11/xx\equiv \dfrac{1}{1/x} and bring that new factor of 1x\dfrac{1}{x} 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 (x2+3x)12x(1+3x)12\left(x^2+3x\right)^{\frac{1}{2}} \equiv x\left(1 + \dfrac{3}{x}\right)^{\frac{1}{2}}, which is valid for x>3|x|>3 and hence will do the trick.


If 1<3x<1 \displaystyle -1 < \frac {3}{x} < 1 then

3<x<3 \displaystyle -3 < x < 3 so how did you get

x>3 \displaystyle |x| > 3 ?
Reply 8
Original post by member910132
If 1<3x<1 \displaystyle -1 < \frac {3}{x} < 1 then

3<x<3 \displaystyle -3 < x < 3 so how did you get

x>3 \displaystyle |x| > 3 ?


When taking the reciprocal, you have to reverse the inequality signs.
Reply 9
Original post by f1mad
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

x=1y \displaystyle x = \frac {1}{y}

even after that, we get

Unparseable latex formula:

\displaystyle \left( \frac {1}{y^2} + \frac {3}{y})^\frac{1}{2}



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

(13y)12×(1+3y)12=(1y2)12×(1+3y)12 \displaystyle (\frac {1}{3y})^\frac {1}{2} \times (1+ \frac{3}{y})^\frac{1}{2} = (\frac{1}{y^2})^\frac{1}{2} \times (1+3y)^\frac{1}{2}

and you will only get the correct answer of 32 \displaystyle \frac{3}{2} 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 x=1y \displaystyle x = \frac{1}{y} ? 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 x \displaystyle x \to \infty then the expansions only valid for 1<x<1 \displaystyle -1< x<1 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 x11/xx\equiv \dfrac{1}{1/x} and bring that new factor of 1x\dfrac{1}{x} 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 (x2+3x)12x(1+3x)12\left(x^2+3x\right)^{\frac{1}{2}} \equiv x\left(1 + \dfrac{3}{x}\right)^{\frac{1}{2}}, which is valid for x>3|x|>3 and hence will do the trick.


Could you go into the "Note that x11/xx\equiv \dfrac{1}{1/x} and bring that new factor of 1x\dfrac{1}{x} into the square root in the denominator." in more detail, I don't really understand ?
(edited 11 years ago)
Reply 10
Original post by f1mad
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: 1<3x<1 -1 < \frac {3}{x}<1

1>x3>1-1>\frac{x}{3}>1 but how can something be smaller than -1 and bigger than 1 ?
(edited 11 years ago)
Original post by member910132
If 1<3x<1 \displaystyle -1 < \frac {3}{x} < 1 then

3<x<3 \displaystyle -3 < x < 3 so how did you get

x>3 \displaystyle |x| > 3 ?

3x<1    x3>1\left|\dfrac{3}{x}\right|<1 \implies \left|\dfrac{x}{3}\right| >1

Original post by member910132
Could you go into the "Note that x11/xx\equiv \dfrac{1}{1/x} and bring that new factor of 1x\dfrac{1}{x} into the square root in the denominator." in more detail, I don't really understand ?

So we have 3xx2+3x+x31x(x2+3x+x)31+3x+1\dfrac{3x}{\sqrt{x^2+3x} +x} \equiv \dfrac{3}{\frac{1}{x}(\sqrt{x^2+3x} +x)} \equiv \dfrac{3}{\sqrt{1 + \frac{3}{x}} + 1}.

Can you take the limit now?
Reply 12
Original post by Farhan.Hanif93
3x<1    x3>1\left|\dfrac{3}{x}\right|<1 \implies \left|\dfrac{x}{3}\right| >1




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

And how can x3 \frac{x}{3} be smaller than -1 and grater than 1?
(edited 11 years ago)
Reply 13
Original post by Farhan.Hanif93


If you wanted to use binomial expansion from the start, you could have noted (x2+3x)12x(1+3x)12\left(x^2+3x\right)^{\frac{1}{2}} \equiv x\left(1 + \dfrac{3}{x}\right)^{\frac{1}{2}}, which is valid for x>3|x|>3 and hence will do the trick.


Leaving aside this confusion of mine about the range of validity,

after expanding it all I get to:

32x98x2 \frac{3}{2x} - \frac{9}{8x^2} and that doesn't give me the answer of 3/2 as x -> infty.
Reply 14
Original post by Farhan.Hanif93
3x<1    x3>1\left|\dfrac{3}{x}\right|<1 \implies \left|\dfrac{x}{3}\right| >1


So we have 3xx2+3x+x31x(x2+3x+x)31+3x+1\dfrac{3x}{\sqrt{x^2+3x} +x} \equiv \dfrac{3}{\frac{1}{x}(\sqrt{x^2+3x} +x)} \equiv \dfrac{3}{\sqrt{1 + \frac{3}{x}} + 1}.

Can you take the limit now?


Yea I got it form here, that is clever thinking though !!
Original post by member910132
Leaving aside this confusion of mine about the range of validity,

after expanding it all I get to:

32x98x2 \frac{3}{2x} - \frac{9}{8x^2} 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.
Reply 16
Right, and to save me from making a brand new thread, can someone address the issue of flipping the inequality sign when taking reciprocals ?

1<x<1    1>1x>1 -1<x<1 \implies -1> \dfrac{1}{x} >1 but how can 1/x be bigger then 1 and smaller than -1 ?

Or is it like this

1<x<1    1>1x and 1x>1 -1<x<1 \implies -1 > \frac{1}{x}\ \text{and} \ \frac{1}{x}>1 ?

Edit: Would x>3 be written as 1>1x and 1x>1 |x| > 3 \ \text{be written as}\ -1 > \frac{1}{x}\ \text{and} \ \frac{1}{x}>1 and cannot it not be written as one inequality ?
(edited 11 years ago)

Quick Reply

Latest