Calculus 3

$\displaystyle\lim_{x \to \infty} x^3\left( \sqrt{x^2+\sqrt{1+x^4}} - x\sqrt{2} \right) \\ \displaystyle\lim_{x \to \infty} x^4\left( \sqrt{1+\sqrt{1+\dfrac1{x^4}}} - \sqrt{2} \right) \\ \displaystyle\lim_{x \to \infty} x^4\left( \sqrt{1+1+\dfrac1{2x^4}+\cdots} - \sqrt{2} \right) \\ \displaystyle\lim_{x \to \infty} \sqrt2x^4\left( \sqrt{1+\left(\dfrac1{4x^4}+\cdots\right)} - 1 \right) \\ \displaystyle\lim_{x \to \infty} \sqrt2x^4\left( \cancel{1}+\dfrac1{8x^4}+\cdots - \cancel{1} \right) \\ \displaystyle\lim_{x \to \infty} \sqrt2x^4\left( \dfrac1{8x^4}+\cdots \right) \implies \dfrac{1}{4\sqrt2} = \boxed{0.178}$ \

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$\text{-: Alternate :- }$

$\displaystyle\lim_{x \to \infty} x^3\left( \sqrt{x^2+\sqrt{1+x^4}} - x\sqrt{2} \right) \cdot \dfrac{\sqrt{x^2+\sqrt{1+x^4}} + x\sqrt{2}}{\sqrt{x^2+\sqrt{1+x^4}} +x\sqrt{2}} \\ \displaystyle\lim_{x \to \infty} x^3\dfrac{\left( x^2+\sqrt{1+x^4} - 2x^2 \right)}{ \left( \sqrt{x^2+\sqrt{1+x^4}} + x\sqrt{2} \right)}\cdot \dfrac{\left(\sqrt{1+x^4} +x^2 \right)}{\left(\sqrt{1+x^4} + x^2 \right)} \\ \displaystyle\lim_{x \to \infty} x^3\dfrac{\left(1+\cancel{x^4} -\cancel{ x^4 }\right)}{ \left( \sqrt{x^2+\sqrt{1+x^4}} + x\sqrt{2} \right)\left(\sqrt{1+x^4} + x^2 \right)} \\ \displaystyle\lim_{x \to \infty} \dfrac{x^3}{ \left( \sqrt{x^2+\sqrt{1+x^4}} + x\sqrt{2} \right)\left(\sqrt{1+x^4} + x^2 \right)} \\ \displaystyle\lim_{x \to \infty} \dfrac{\cancel{x^3}}{ \cancel{x^3}\left( \sqrt{1+\sqrt{1+\dfrac1{x^4}}} + \sqrt{2} \right)\left(\sqrt{1+\dfrac1{x^4}} + 1 \right)} \\ = \dfrac{1}{4\sqrt2} \implies \boxed{0.178}$