The Student Room Group

Limits

Ok so need help with this question. Sadly it's in the textbook so all I have is the answer.
The problem I have is I don't see how the x x terms cancel.

limx[(x2+3x)12x] \displaystyle\lim_{x\to \infty}[ (x^2+3x)^\frac{1}{2} -x ]
Any help would be appreciated.
If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.
Original post by Rainingshame
Ok so need help with this question. Sadly it's in the textbook so all I have is the answer.
The problem I have is I don't see how the x x terms cancel.

limx[(x2+3x)12x] \displaystyle\lim_{x\to \infty}[ (x^2+3x)^\frac{1}{2} -x ]
Any help would be appreciated.
If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.


Do you need to prove the limit or just find it.

If it is the latter just try some big numbers for x such as 1000000.
Reply 2
Original post by Rainingshame
Ok so need help with this question. Sadly it's in the textbook so all I have is the answer.
The problem I have is I don't see how the x x terms cancel.

limx[(x2+3x)12x] \displaystyle\lim_{x\to \infty}[ (x^2+3x)^\frac{1}{2} -x ]
Any help would be appreciated.
If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.


Does your textbook give any similar examples? It's difficult to give a hint for this without explaining exactly how to do it if you haven't seen it before.
Original post by 2710
(x2+3x)12(x^2+3x)^\frac{1}{2}

When x gets really big, i.e tends to infinity, the 3x becomes negligible, because it is so tiny compared to the x^2.


And then?
Reply 4
Original post by Mr M
And then?


What's your point?
Reply 5
Original post by davros
Does your textbook give any similar examples? It's difficult to give a hint for this without explaining exactly how to do it if you haven't seen it before.


Nope, all the examples given are of functions divided by another function. That's why this one confused me.
Original post by Rainingshame
Nope, all the examples given are of functions divided by another function. That's why this one confused me.


Can you turn it into something divided by something?
Reply 7
Original post by Mr M
Do you need to prove the limit or just find it.

If it is the latter just try some big numbers for x such as 1000000.


Prove, though we can state the expansion for the binomial, exponential function ln(1+x) ln(1+x) cosine and sine without proof.
Reply 8
Try rationalising
Original post by 2710
What's your point?


Funnily enough that was what I was asking you.
Original post by Rainingshame
Prove, though we can state the expansion for the binomial, exponential function ln(1+x) ln(1+x) cosine and sine without proof.


Ok try multiplying by x2+3x+xx2+3x+x\displaystyle \frac{\sqrt{x^2+3x}+x}{\sqrt{x^2+3x}+x} and then see if you can bring the binomial expansion into play.
Original post by Rainingshame
Ok so need help with this question. Sadly it's in the textbook so all I have is the answer.
The problem I have is I don't see how the x x terms cancel.

limx[(x2+3x)12x] \displaystyle\lim_{x\to \infty}[ (x^2+3x)^\frac{1}{2} -x ]
Any help would be appreciated.
If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.

[br]x2+3xx=((x+32)294)1/2x[br]\displaystyle[br]\sqrt{x^2 + 3x} - x = ((x+\frac{3}{2})^2 - \frac{9}{4})^{1/2} - x[br]
For large x, the 94\displaystyle - \frac{9}{4} is insignificant thus the equation becomes:
((x+32)2)12x=x+32x=32\displaystyle ((x+\frac{3}{2})^2)^{\frac{1}{2}} - x = x + \frac{3}{2} - x = \frac{3}{2} Thus the answer is 1.5
(edited 10 years ago)
Original post by Mr M
Ok try multiplying by x2+3x+xx2+3x+x\displaystyle \frac{\sqrt{x^2+3x}+x}{\sqrt{x^2+3x}+x} and then see if you can bring the binomial expansion into play.


I have to separate the x3+3x \sqrt{x^3+3x} into 3x1+x3 \sqrt{3x}\sqrt{1+\frac{x}{3}} I think. Let me see what I get once I expand that.
Original post by Rainingshame
I have to separate the x2+3x \sqrt{x^2+3x} into x1+3x x\sqrt{1+\frac{3}{x}} I think. Let me see what I get once I expand that.


Fixed that for you.
Original post by Mr M
Fixed that for you.


Thanks! I didn't know you could expand it in that form though, it makes it more obvious why you can find the limit though, and makes the answer downright obvious.
Reply 15
Original post by Rainingshame
Ok so need help with this question. Sadly it's in the textbook so all I have is the answer.
The problem I have is I don't see how the x x terms cancel.

limx[(x2+3x)12x] \displaystyle\lim_{x\to \infty}[ (x^2+3x)^\frac{1}{2} -x ]
Any help would be appreciated.
If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.


Now you've seen a few suggestions, here is the cleanest way to attack this.

As MrM says, multiply what you've got by:

x2+3x+xx2+3x+x\displaystyle \frac{\sqrt{x^2+3x}+x}{\sqrt{x^2+3x}+x}

You then end up with:

3xx2+3x+x\displaystyle \frac{3x}{\sqrt{x^2+3x}+x}
or
31+(3/x)+1\displaystyle \frac{3}{\sqrt{1+(3/x)}+1}

You can now apply the algebra of limits directly, as x>x->\infty, 3/x->0, 1+(3/x)->1, the square root tends to 1 and the denominator tends to 2.

Hence the limit is 3/2.

This method avoids dealing with infinite expansions or using any claims about terms being "negligible" without proof :smile:
Original post by davros
Now you've seen a few suggestions, here is the cleanest way to attack this.

As MrM says, multiply what you've got by:

x2+3x+xx2+3x+x\displaystyle \frac{\sqrt{x^2+3x}+x}{\sqrt{x^2+3x}+x}

You then end up with:

3xx2+3x+x\displaystyle \frac{3x}{\sqrt{x^2+3x}+x}
or
31+(3/x)+1\displaystyle \frac{3}{\sqrt{1+(3/x)}+1}

You can now apply the algebra of limits directly, as x>x->\infty, 3/x->0, 1+(3/x)->1, the square root tends to 1 and the denominator tends to 2.

Hence the limit is 3/2.

This method avoids dealing with infinite expansions or using any claims about terms being "negligible" without proof :smile:


I tend to avoid the negligible ones anyway. Not really sure when you're allowed to use it so.... I still used the infinite expansion. It's a bit of extra writing but that's what they're expecting you to do so I though, Why not?
Reply 17
Original post by Rainingshame
I tend to avoid the negligible ones anyway. Not really sure when you're allowed to use it so.... I still used the infinite expansion. It's a bit of extra writing but that's what they're expecting you to do so I though, Why not?


If that's what they teach you for A level then that's fine - technically you can't assume that you can take an infinite limit inside an infinite series and get the right answer, and the method I provided avoids the need for infinite series altogether, but go with whatever's allowed by your syllabus :smile:

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