Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}$

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

\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}