lim _{x rightarrow infty} x^3 left{sqrt{x^2+s | vedclass

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Question

(B) We evaluate the limit: $\lim _{x \rightarrow \infty} x^3 \left(\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right)$ Rationalizing the expression: $= \lim _{

Vedclass · Accepted Answer

$\frac{1}{4 \sqrt{2}}$

Vedclass · Answer

(B) We evaluate the limit: $\lim _{x \rightarrow \infty} x^3 \left(\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right)$ Rationalizing the expression: $= \lim _{x \rightarrow \infty} \frac{x^3 \left(x^2+\sqrt{1+x^4}-2 x^2\right)}{\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}} = \lim _{x \rightarrow \infty} \frac{x^3 \left(\sqrt{1+x^4}-x^2\right)}{\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}}$ Rationalizing the numerator again: $= \lim _{x \rightarrow \infty} \frac{x^3 \left(1+x^4-x^4\right)}{\left(\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}\right) \left(\sqrt{1+x^4}+x^2\right)}$ Dividing by $x^3$ in the denominator: $= \lim _{x \rightarrow \infty} \frac{x^3}{x \left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}}+\sqrt{2}\right) \cdot x^2 \left(\sqrt{\frac{1}{x^4}+1}+1\right)}$ $= \lim _{x \rightarrow \infty} \frac{1}{\left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}}+\sqrt{2}\right) \left(\sqrt{\frac{1}{x^4}+1}+1\right)}$ Substituting $x \rightarrow \infty$: $= \frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)} = \frac{1}{(\sqrt{2}+\sqrt{2})(2)} = \frac{1}{2 \sqrt{2} \cdot 2} = \frac{1}{4 \sqrt{2}}$

$\lim _{x \rightarrow \infty} x^3 \left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\} = $

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