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

To find the limit of the given expression as x approaches 0, we can use L'Hopital's rule.<br /><br />L'Hopital's rule states that if the limit of a function as x approaches a certain value is in the form of 0/0 or ∞/∞, then the limit can be found by taking the derivative of the numerator and denominator separately and then taking the limit of the ratio of the derivatives.<br /><br />In this case, the given expression is in the form of 0/0 as x approaches 0. Therefore, we can apply L'Hopital's rule.<br /><br />Taking the derivative of the numerator, we get:<br />$\frac{d}{dx}(\sqrt[4]{x^4+1} - \sqrt{x^2+1}) = \frac{1}{4}(x^4+1)^{-3/4} \cdot 4x^3 - \frac{1}{2}(x^2+1)^{-1/2} \cdot 2x$<br /><br />Simplifying, we get:<br />$\frac{d}{dx}(\sqrt[4]{x^4+1} - \sqrt{x^2+1}) = x^3(x^4+1)^{-3/4} - x(x^2+1)^{-1/2}$<br /><br />Taking the derivative of the denominator, we get:<br />$\frac{d}{dx}(x^2) = 2x$<br /><br />Now, we can apply L'Hopital's rule to find the limit:<br />$\lim_{x\to 0} \frac{\sqrt[4]{x^4+1} - \sqrt{x^2+1}}{x^2} = \lim_{x\to 0} \frac{x^3(x^4+1)^{-3/4} - x(x^2+1)^{-1/2}}{2x}$<br /><br />Simplifying, we get:<br />$\lim_{x\to 0} \frac{x^3(x^4+1)^{-3/4} - x(x^2+1)^{-1/2}}{2x} = \lim_{x\to 0} \frac{x^2(x^4+1)^{-3/4} - (x^2+1)^{-1/2}}{2}$<br /><br />As x approaches 0, the numerator approaches 0, and the denominator approaches 2. Therefore, the limit is 0.<br /><br />So, the answer is 0.