# 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).

• Do you know l'Hopital? – wythagoras May 6 '15 at 19:21
• I forgot to include that I still hasn't studied l'Hopital. – BlackBrain May 6 '15 at 19:27
• The key (as in the answers) is to rationalize the "interesting" part, i.e. the indeterminate parts. There are two of these (both numerator and denominator are indeterminate), so you should rationalize both. – aes May 6 '15 at 19:30

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.

• +$1$ Good answer! You beat me to it :) – Clayton May 6 '15 at 19:29
• @Clayton Thanks. Yet someone didn't like it. Oh, well. – Timbuc May 6 '15 at 19:29

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.

Just multiply with $\frac{\sqrt{x-1}+\sqrt{x-2}}{\sqrt{x-1}+\sqrt{x-2}}$ aswell. Then it should be easy to evaluate.

\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}