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

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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.

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.