lim _(xarrow 2)(sqrt (2x-3)-1)/(x-2)=

To evaluate the limit \(\lim_{x \to 2} \frac{\sqrt{2x - 3} - 1}{x - 2}\), we can use algebraic manipulation and L'Hôpital's Rule.<br /><br />First, let's rewrite the expression:<br />\[<br />\lim_{x \to 2} \frac{\sqrt{2x - 3} - 1}{x - 2}<br />\]<br /><br />We notice that direct substitution of \(x = 2\) results in an indeterminate form \(\frac{0}{0}\). To resolve this, we can multiply the numerator and the denominator by the conjugate of the numerator:<br />\[<br />\lim_{x \to 2} \frac{(\sqrt{2x - 3} - 1)(\sqrt{2x - 3} + 1)}{(x - 2)(\sqrt{2x - 3} + 1)}<br />\]<br /><br />The numerator simplifies as follows:<br />\[<br />(\sqrt{2x - 3} - 1)(\sqrt{2x - 3} + 1) = (2x - 3) - 1^2 = 2x - 4<br />\]<br /><br />So the expression becomes:<br />\[<br />\lim_{x \to 2} \frac{2x - 4}{(x - 2)(\sqrt{2x - 3} + 1)}<br />\]<br /><br />We can factor the numerator:<br />\[<br />2x - 4 = 2(x - 2)<br />\]<br /><br />Thus, the expression simplifies to:<br />\[<br />\lim_{x \to 2} \frac{2(x - 2)}{(x - 2)(\sqrt{2x - 3} + 1)} = \lim_{x \to 2} \frac{2}{\sqrt{2x - 3} + 1}<br />\]<br /><br />Now, we can substitute \(x = 2\) into the simplified expression:<br />\[<br />\frac{2}{\sqrt{2 \cdot 2 - 3} + 1} = \frac{2}{\sqrt{4 - 3} + 1} = \frac{2}{\sqrt{1} + 1} = \frac{2}{1 + 1} = \frac{2}{2} = 1<br />\]<br /><br />Therefore, the limit is:<br />\[<br />\boxed{1}<br />\]