Bbluncnutb npenen lim _(xarrow 10)(-sqrt (10)+sqrt (x))/(x-10)

To evaluate the limit \(\lim_{x \to 10} \frac{-\sqrt{10} + \sqrt{x}}{x - 10}\), we can use algebraic manipulation and L'Hôpital's Rule.<br /><br />First, let's rewrite the expression:<br /><br />\[<br />\lim_{x \to 10} \frac{-\sqrt{10} + \sqrt{x}}{x - 10}<br />\]<br /><br />As \(x\) approaches 10, both the numerator and the denominator approach 0, which means we can apply L'Hôpital's Rule. L'Hôpital's Rule states that if \(\lim_{x \to c} \frac{f(x)}{g(x)}\) results in an indeterminate form \(\frac{ or \(\frac{\infty}{\infty}\), then:<br /><br />\[<br />\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}<br />\]<br /><br />provided the limit on the right-hand side exists.<br /><br />Let \(f(x) = -\sqrt{10} + \sqrt{x}\) and \(g(x) = x - 10\).<br /><br />First, we find the derivatives of \(f(x)\) and \(g(x)\):<br /><br />\[<br />f'(x) = \frac{d}{dx}(-\sqrt{10} + \sqrt{x}) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}<br />\]<br /><br />\[<br />g'(x) = \frac{d}{dx}(x - 10) = 1<br />\]<br /><br />Now, we apply L'Hôpital's Rule:<br /><br />\[<br />\lim_{x \to 10} \frac{-\sqrt{10} + \sqrt{x}}{x - 10} = \lim_{x \to 10} \frac{f'(x)}{g'(x)} = \lim_{x \to 10} \frac{\frac{1}{2\sqrt{x}}}{1} = \lim_{x \to 10} \frac{1}{2\sqrt{x}}<br />\]<br /><br />As \(x\) approaches 10:<br /><br />\[<br />\lim_{x \to 10} \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{10}}<br />\]<br /><br />Therefore, the limit is:<br /><br />\[<br />\lim_{x \to 10} \frac{-\sqrt{10} + \sqrt{x}}{x - 10} = \frac{1}{2\sqrt{10}}<br />\]<br /><br />This can also be written as:<br /><br />\[<br />\lim_{x \to 10} \frac{-\sqrt{10} + \sqrt{x}}{x - 10} = \frac{\sqrt{10}}{20}<br />\]<br /><br />So, the final\lim_{x \frac{-\sqrt{10} + \sqrt{x}}{x - 10} = \frac{\sqrt{10}}{20}<br />\]