Square roots and limits.

[steely]

Can't quite figure this out...

 

Why does sqrt[x(x+1)] tend to x+1/2 and not just x as x becomes very large?

[DJBruce]

Maybe I am misunderstanding you but I think you want to evaluate the limit:

 

[math]\lim_{x\to \infty} \sqrt{x^{2}+x}[/math]

 

You are wondering the why:

[math]\lim_{x\to \infty} \sqrt{x^{2}+x}= \infty + \frac{1}{2}[/math]

 

not: [math]\lim_{x\to \infty} \sqrt{x^{2}+x}= \infty [/math]

 

My questions for you are. What is [math]\frac{1}{2}[/math] bigger than infinity? Can you be bigger than infinity?

Edited November 30, 2009 by Cap'n Refsmmat
fixed your LaTeX for you :-p

[ajb]

You could consider a Taylor expansion in the neighbourhood of infinity. In which case you would get [math] f(x) \approx x + \frac{1}{2}[/math].

 

The limit of this as [math]x \rightarrow \infty[/math] would then be [math]\infty[/math].

[D H]

Can't quite figure this out...

 

Why does sqrt[x(x+1)] tend to x+1/2 and not just x as x becomes very large?

 

[math]\sqrt{x(x+1)} = \sqrt{x^2+x} = \sqrt{x^2\left(1+\frac 1 x\right)}

= |x|\,\sqrt{1+\frac 1 x}[/math]

 

Assuming x is positive, this becomes

 

[math]\sqrt{x(x+1)} = x\sqrt{1+\frac 1 x}[/math]

 

This is still exact. For small numbers [math]|a| \ll 1,\, \sqrt{1+a} \approx 1+\frac 1 2 a[/math]. Using this to approximate the radical in the above for large x,

 

[math]\sqrt{1+\frac 1 x} \approx 1 + \frac 1 {2x},\,|x| \ggg 1 [/math]

 

Thus for large x,

 

[math]\sqrt{x(x+1)} = x\sqrt{1+\frac 1 x}

\approx x\left(1 + \frac 1 {2x}\right) =x + \frac 1 2[/math]

[timo]

Or you just claim that with saying [math] \sqrt{x(x+1)} = \sqrt{x^2+x} \approx \sqrt{x^2} = x[/math] you throw away a (large) term x under the root while with [math] \sqrt{x(x+1)} = \sqrt{x^2 + x} = \sqrt{ (x+1/2)^2 - 1/4} \approx \sqrt{ (x+1/2)^2} = x+1/2 [/math] you only throw away a small constant so the latter approximation is better.

[D H]

As a sanity check, some numbers:

 

[math]\aligned

x \quad & \,\,\sqrt{x^2+x} \\

\phantom{0000}1 \quad & \phantom{0000}1.4142 \\

\phantom{000}10 \quad & \phantom{000}10.4881 \\

\phantom{00}100 \quad & \phantom{00}100.4988 \\

\phantom{0}1000 \quad & \phantom{0}1000.4998 \\

10000 \quad & 10000.49999

\endaligned[/math]