Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}$ I'm trying to solve evaluate this limit
$$\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}.$$
I've tried to rationalize the denominator but this is what I've got
$$\lim_{x\to\infty}(\sqrt{x-1} - \sqrt{x-2})({\sqrt{x-2} + \sqrt{x-3}})$$
and I don't know how to remove these indeterminate forms $(\infty - \infty)$.
EDIT: without l'Hospital's rule (if possible).
 A: Note that
$$\sqrt{x-1}-\sqrt{x-2} = \dfrac1{\sqrt{x-1}+\sqrt{x-2}}$$
and
$$\dfrac1{\sqrt{x-2}-\sqrt{x-3}} = \sqrt{x-2}+\sqrt{x-3}$$
We hence have
$$\dfrac{\sqrt{x-1}-\sqrt{x-2}}{\sqrt{x-2}-\sqrt{x-3}} = \dfrac{\sqrt{x-2}+\sqrt{x-3}}{\sqrt{x-1}+\sqrt{x-2}}$$
We have
$$\dfrac{\sqrt{x-3}}{\sqrt{x-2}}<\dfrac{\sqrt{x-2}+\sqrt{x-3}}{\sqrt{x-1}+\sqrt{x-2}} < 1$$
Now conclude what you want.
A: Just multiply with $\frac{\sqrt{x-1}+\sqrt{x-2}}{\sqrt{x-1}+\sqrt{x-2}}$ aswell. Then it should be easy to evaluate.
A: \begin{align}
\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}
&=\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}\,\,
\frac{\sqrt{x-1} + \sqrt{x-2}}{\sqrt{x-1} + \sqrt{x-2}}\,\,
\frac{\sqrt{x-2} + \sqrt{x-3}}{\sqrt{x-2} + \sqrt{x-3}}\\ \ \\
&=\frac{x-1-(x-2)}{x-2-(x-3)}\,\frac{\sqrt{x-2} + \sqrt{x-3}}{\sqrt{x-1} + \sqrt{x-2}}=\frac{\sqrt{x-2} + \sqrt{x-3}}{\sqrt{x-1} + \sqrt{x-2}}\\ \ \\
&=\frac{\sqrt{1-2/x}+\sqrt{1-3/x}}{\sqrt{1-1/x}+\sqrt{1-2/x}}\to\frac22=1
\end{align}
A: Fill in details:
As $\;x\to\infty\;$ we can assume $\;x>0\;$ , so:
$$\frac{\sqrt{x-1}-\sqrt{x-2}}{\sqrt{x-2}-\sqrt{x-3}}=\frac{\sqrt{x-2}+\sqrt{x-3}}{\sqrt{x-1}+\sqrt{x-2}}=\frac{\sqrt{1-\frac2x}+\sqrt{1-\frac3x}}{\sqrt{1-\frac1x}+\sqrt{1-\frac2x}}\xrightarrow[x\to\infty]{}1$$
Further hint: the first step was multiplying by conjugate of both the numerator and the denominator.
