Let f(x)=sqrt (x)-5 Then lim _(harrow 0)(f(2+h)-f(2))/(h)=

To find the limit \(\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}\), we first need to determine the function \(f(x)\) and then evaluate the difference quotient.<br /><br />Given \(f(x) = \sqrt{x} - 5\), we can find \(f(2)\) and \(f(2+h)\):<br /><br />\[ f(2) = \sqrt{2} - 5 \]<br />\[ f(2+h) = \sqrt{2+h 5 \]<br /><br />Now, we can substitute these values into the difference quotient:<br /><br />\[ \frac{f(2+h) - f(2)}{h} = \frac{(\sqrt{2+h} - 5) - (\sqrt{2} - 5)}{h} \]<br /><br />Simplifying the numerator, we have:<br /><br />\[ \frac{\sqrt{2+h} - \sqrt{2}}{h} \]<br /><br />To evaluate this limit, we can use the conjugate method. Multiply the numerator and denominator by the conjugate of the numerator:<br /><br />\[ \frac{\sqrt{2+h} - \sqrt{2}}{h} \cdot \frac{\sqrt{2+h} + \sqrt{2}}{\sqrt{2+h} + \sqrt{2}} \]<br /><br />This simplifies to:<br /><br />\[ \frac{(\sqrt{2+h} - \sqrt{2})(\sqrt{2+h} + \sqrt{2})}{h(\sqrt{2+h} + \sqrt{2})} = \frac{(2+h) - 2}{h(\sqrt{2+h} + \sqrt{2})} = \frac{h}{h(\sqrt{2+h} + \sqrt{2})} = \frac{1}{\sqrt{2+h} + \sqrt{2}} \]<br /><br />As \(h\) approaches 0, \(\sqrt{2+h}\) approaches \(\sqrt{2}\). Therefore, the limit is:<br /><br />\[ \lim_{h \to 0} \frac{1}{\sqrt{2+h} + \sqrt{2}} = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \]<br /><br />So, the correct answer is \(\frac{\sqrt{2}}{4}\).