lim _(xarrow 3)(sqrt (2x+3)-x)/(3-x)

To evaluate the limit \(\lim_{x \to 3} \frac{\sqrt{2x+3} - x}{3 - x}\), we can use algebraic manipulation and L'Hôpital's Rule.<br /><br />First, let's rewrite the expression:<br />\[<br />\lim_{x \to 3} \frac{\sqrt{2x+3} - x}{3 - x}<br />\]<br /><br />We notice that direct substitution of \(x = 3\) results in an indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states that if \(\lim_{x \to c} \frac{f(x)}{g(x)}\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:<br />\[<br />\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}<br />\]<br /><br />First, we need to find the derivatives of the numerator and the denominator.<br /><br />For the numerator \( \sqrt{2x+3} - x \):<br />\[<br />\frac{d}{dx} \left( \sqrt{2x+3} - x \right) = \frac{d}{dx} \left( (2x+3)^{1/2} \right) - \frac{d}{dx} (x)<br />\]<br />Using the chain rule:<br />\[<br />\frac{d}{dx} \left( (2x+3)^{1/2} \right) = \frac{1}{2} (2x+3)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x+3}}<br />\]<br />Thus,<br />\[<br />\frac{d}{dx} \left( \sqrt{2x+3} - x \right) = \frac{1}{\sqrt{2x+3}} - 1<br />\]<br /><br />For the denominator \( 3 - x \):<br />\[<br />\frac{d}{dx} (3 - x) = -1<br />\]<br /><br />Now, applying L'Hôpital's Rule:<br />\[<br />\lim_{x \to 3} \frac{\sqrt{2x+3} - x}{3 - x} = \lim_{x \to 3} \frac{\frac{1}{\sqrt{2x+3}} - 1}{-1}<br />\]<br /><br />Simplify the expression inside the limit:<br />\[<br />\lim_{x \to 3} \frac{\frac{1 - \sqrt{2x+3}}{\sqrt{2x+3}}}{-1} = \lim_{x \to 3} \frac{-(1 - \sqrt{2x+3})}{\sqrt{2x+3}}<br />\]<br /><br />Substitute \( x = 3 \):<br />\[<br />\lim_{x \to 3} \frac{-(1 - \sqrt{2 \cdot 3 + 3})}{\sqrt{2 \cdot 3 + 3}} = \frac{-(1 - \sqrt{9})}{\sqrt{9}} = \frac{-(1 - 3)}{3} = \frac{-(-2)}{3} = \frac{2}{3}<br />\]<br /><br />Therefore, the limit is:<br />\[<br />\boxed{\frac{2}{3}}<br />\]