lim _(xarrow 0)(sqrt [4](x^4+1)-sqrt (x^2+1))/(x^2)

To evaluate the limit \(\lim_{x \to 0} \frac{\sqrt[4]{x^4 + 1} - \sqrt{x^2 + 1}}{x^2}\), we can use L'Hôpital's Rule since it is in an indeterminate form \(\frac{0}{0}\).<br /><br />First, let's rewrite the expression:<br />\[ \lim_{x \to 0} \frac{\sqrt[4]{x^4 + 1} - \sqrt{x^2 + 1}}{x^2} \]<br /><br />We need to 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 />\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]<br /><br />First, we find the derivatives of the numerator and the denominator.<br /><br />The numerator is \( \sqrt[4]{x^4 + 1} - \sqrt{x^2 + 1} \).<br /><br />Let \( u = \sqrt[4]{x^4 + 1} \) and \( v = \sqrt{x^2 + 1} \).<br /><br />Then, the derivatives are:<br />\[ u' = \frac{d}{dx} \left( (x^4 + 1)^{1/4} \right) = \frac{1}{4} (x^4 + 1)^{-3/4} \cdot 4x^3 = \frac{3x^3}{(x^4 + 1)^{3/4}} \]<br />\[ v' = \frac{d}{dx} \left( (x^2 + 1)^{1/2} \right) = \frac{1}{2} (x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \]<br /><br />Now, applying L'Hôpital's Rule:<br />\[ \lim_{x \to 0} \frac{\sqrt[4]{x^4 + 1} - \sqrt{x^2 + 1}}{x^2} = \lim_{x \to 0} \frac{\frac{3x^3}{(x^4 + 1)^{3/4}} - \frac{x}{\sqrt{x^2 + 1}}}{2x} \]<br /><br />Simplify the expression:<br />\[ = \lim_{x \to 0} \left( \frac{3x^3}{2x (x^4 + 1)^{3/4}} - \frac{x}{2x \sqrt{x^2 + 1}} \right) \]<br />\[ = \lim_{x \to 0} \left( \frac{3x^2}{2 (x^4 + 1)^{3/4}} - \frac{1}{2 \sqrt{x^2 + 1}} \right) \]<br /><br />As \( x \to 0 \):<br />\[ \frac{3x^2}{2 (x^4 + 1)^{3/4}} \to 0 \]<br />\[ \frac{1}{2 \sqrt{x^2 + 1}} \to \frac{1}{2 \sqrt{1}} = \frac{1}{2} \]<br /><br />Therefore, the limit is:<br />\[ \lim_{x \to 0} \frac{\sqrt[4]{x^4 + 1} - \sqrt{x^2 + 1}}{x^2} = \frac{1}{2} \]