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Limits Help

What is the limit of f(x) as x approaches infinity, where f(x) = [((x^2)+3x)^0.5) - x]

That should read; open bracket, x squared plus 3x, close bracket to the power of half, - x.

I've tried to use a binomial expansion, firstly I took out a factor of (3x)^0.5

Which yielded; (3x)^0.5 [1+(x/3)]^0.5 - x

I had used the binominal expansion 1+nx+n(n-1)x^2/2!+.... where n<1 however I cannot manage to obtain the correct answer of 3/2.

Help would be appreciated
Reply 1
Avatar for Dalilsp
Dalilsp
9
Try rewriting f(x) by rationalising it
Reply 2
Avatar for ArcRaman
ArcRaman
OP
2
Original post by Dalilsp
Try rewriting f(x) by rationalising it


Ok, I just tried that now...

3x/[(x^2+3x)^0.5+x]

Numerator: 3
Denominator: Sqrt(3/x)[expansion terms] + 1

As x--->infinity.

Denominator tends to 1? Yielding a limit of 3?

Wrong though, I've made a mistake somewhere
Reply 3
Avatar for ztibor
ztibor
10
Original post by ArcRaman
Ok, I just tried that now...

3x/[(x^2+3x)^0.5+x]

Numerator: 3
Denominator: Sqrt(3/x)[expansion terms] + 1

As x--->infinity.

Denominator tends to 1? Yielding a limit of 3?

Wrong though, I've made a mistake somewhere



in the denominator

x2+3x+x=x(1+3x+1)\displaystyle \sqrt{x^2+3x}+x=x\left (\sqrt{1+\frac{3}{x}}+1\right )

x factor will be cancelled

and as x-> infty the denominator

1+3∞+1=1+0+1=2\displaystyle \sqrt{1+\frac{3}{\infty}}+1 = \sqrt{1+0} +1=2
(edited 9 years ago)
Reply 4
Avatar for ArcRaman
ArcRaman
OP
2
Original post by ztibor
in the denominator

x2+3x+x=x(1+3x+1)\displaystyle \sqrt{x^2+3x}+x=x\left (\sqrt{1+\frac{3}{x}}+1\right )

x factor will be cancelled

and as x-> infty the denominator

1+3∞+1=1+0+1=2\displaystyle \sqrt{1+\frac{3}{\infty}}+1 = \sqrt{1+0} +1=2


Thanks man

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