lim [sqrt(1+x)-sqrt(x-1)]

x->infinity

i tried to move it to the denominator like this:

lim 1/[sqrt(x+1)+sqrt(x-1)]

but here im stuck, i think it should be zero, but hoe do i prove it?

thanks in advance.

Printable View

- Jul 4th 2006, 01:30 AMDangerManneed help on calculating a limit.
lim [sqrt(1+x)-sqrt(x-1)]

x->infinity

i tried to move it to the denominator like this:

lim 1/[sqrt(x+1)+sqrt(x-1)]

but here im stuck, i think it should be zero, but hoe do i prove it?

thanks in advance. - Jul 4th 2006, 03:13 AMtopsquarkQuote:

Originally Posted by**DangerMan**

$\displaystyle \lim_{x \to \infty} \left ( \sqrt{1+x} - \sqrt{x-1} \right ) \approx \lim_{x \to \infty} \left ( \sqrt{x} - \sqrt{x} \right ) \to 0$

-Dan - Jul 4th 2006, 03:25 AMDangerMan
but i asked for a proof, do you know of a rigorous way to prove it?

thanks. - Jul 4th 2006, 03:34 AMTD!
Depends on "how much rigour" you want...

You could always go back to the definition, but I wouldn't do that unless necessary.

Multiply numerator and denominator with the complement of the numerator to get:

$\displaystyle

\frac{{\left( {\sqrt {x + 1} - \sqrt {x - 1} } \right)\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}}{{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}} = \frac{2}{{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}}

$

Now the numerator is constant and positive and the denominator clearly goes to +infinity, yielding 0 as limit. - Jul 4th 2006, 06:49 AMThePerfectHacker
You can do even this,

note that for $\displaystyle x\geq 1$ we have,

$\displaystyle 0\leq \sqrt{x+1}-\sqrt{x-1} \leq \frac{1}{\sqrt{.5x}}$

But,

$\displaystyle \lim_{x\to\infty} 0=0$ and,

$\displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{.5 x}}=0$

Thus, you have a function,

$\displaystyle f(x)=\sqrt{x+1}-\sqrt{x-1}$

squeezed between two functions which have the same limit. Conclude from the squeeze theorem that,

$\displaystyle \lim_{x\to\infty}\sqrt{x+1}-\sqrt{x-1}=0$