Evaluate the limit if it exists. If not, enter DNE. lim _(xarrow 1)(sqrt (x)-x^2)/(1-sqrt (x))=

To evaluate the limit \(\lim_{x \to 1} \frac{\sqrt{x} - x^2}{1 - \sqrt{x}}\), we can start by substituting \( x = 1 \) into the expression to see if it leads to an indeterminate form.<br /><br />\[<br />\frac{\sqrt{1} - 1^2}{1 - \sqrt{1}} = \frac{0}{0}<br />\]<br /><br />Since this is an indeterminate form, we need to manipulate the expression algebraically to resolve it. We can use the technique of multiplying by the conjugate of the numerator or denominator to simplify the expression.<br /><br />Let's multiply the numerator and the denominator by the conjugate of the numerator, which is \(\sqrt{x} + x^2\):<br /><br />\[<br />\lim_{x \to 1} \frac{(\sqrt{x} - x^2)(\sqrt{x} + x^2)}{(1 - \sqrt{x})(\sqrt{x} + x^2)}<br />\]<br /><br />The numerator simplifies as follows:<br /><br />\[<br />(\sqrt{x} - x^2)(\sqrt{x} + x^2) = (\sqrt{x})^2 - (x^2)^2 = x - x^4<br />\]<br /><br />So the expression becomes:<br /><br />\[<br />\lim_{x \to 1} \frac{x - x^4}{(1 - \sqrt{x})(\sqrt{x} + x^2)}<br />\]<br /><br />Next, we simplify the denominator:<br /><br />\[<br />(1 - \sqrt{x})(\sqrt{x} + x^2) = 1 \cdot \sqrt{x} + 1 \cdot x^2 - \sqrt{x} \cdot \sqrt{x} - \sqrt{x} \cdot x^2 = \sqrt{x} + x^2 - x - x^2\sqrt{x}<br />\]<br /><br />Simplifying further:<br /><br />\[<br />= 1 - x - x^2<br />\]<br /><br />So the limit now is:<br /><br />\[<br />\lim_{x \to 1} \frac{x - x^4}{1 - x - x^2}<br />\]<br /><br />We can factor the numerator as:<br /><br />\[<br />x - x^4 = x(1 - x^3)<br />\]<br /><br />Thus, the limit becomes:<br /><br />\[<br />\lim_{x \to 1} \frac{x(1 - x^3)}{1 - x - x^2}<br />\]<br /><br />As \( x \to 1 \), both the numerator and the denominator approach 0, so we can use L'Hôpital's Rule, which states that if \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}\), then \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\).<br /><br />Differentiating the numerator and the denominator with respect to \( x \):<br /><br />\[<br />\frac{d}{dx}[x(1 - x^3)] = 1 - 3x^2<br />\]<br /><br />\[<br />\frac{d}{dx}[1 - x - x^2] = -1 - 2x<br />\]<br /><br />So the limit becomes:<br /><br />\[<br />\lim_{x \to 1} \frac{1 - 3x^2}{-1 - 2x}<br />\]<br /><br />Substituting \( x = 1 \):<br /><br />\[<br />\frac{1 - 3(1)^2}{-1 - 2(1)} = \frac{1 - 3}{-1 - 2} = \frac{-2}{-3} = \frac{2}{3}<br />\]<br /><br />Therefore, the limit is:<br /><br />\[<br />\boxed{\frac{2}{3}}<br />\]