# Thread:

1. ## need 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.

2. Originally Posted by DangerMan
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.
A simple (nonrigorous) way to do it is to look at $\sqrt{x \pm 1}$. When x is large, the 1 is so small as to be ignorable. Thus
$\lim_{x \to \infty} \left ( \sqrt{1+x} - \sqrt{x-1} \right ) \approx \lim_{x \to \infty} \left ( \sqrt{x} - \sqrt{x} \right ) \to 0$

-Dan

3. but i asked for a proof, do you know of a rigorous way to prove it?

thanks.

4. 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:

$<br /> \frac{{\left( {\sqrt {x + 1} - \sqrt {x - 1} } \right)\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}}{{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}} = \frac{2}{{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}}<br />$

Now the numerator is constant and positive and the denominator clearly goes to +infinity, yielding 0 as limit.

5. You can do even this,
note that for $x\geq 1$ we have,
$0\leq \sqrt{x+1}-\sqrt{x-1} \leq \frac{1}{\sqrt{.5x}}$
But,
$\lim_{x\to\infty} 0=0$ and,
$\lim_{x\to\infty} \frac{1}{\sqrt{.5 x}}=0$

Thus, you have a function,
$f(x)=\sqrt{x+1}-\sqrt{x-1}$
squeezed between two functions which have the same limit. Conclude from the squeeze theorem that,
$\lim_{x\to\infty}\sqrt{x+1}-\sqrt{x-1}=0$
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