thnx :)
evalute this limit
\lim_{x\rightarrow \infty }x^{3}\left\{\sqrt{x^{2}+\sqrt{1+x^{4}}}-x\sqrt{2} \right\}
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2 Answers
\\\lim_{x\rightarrow \infty }x^{3}\left\{\sqrt{x^{2}+\sqrt{1+x^{4}}}-x\sqrt{2} \right\} \\=\lim_{x\rightarrow \infty }x^{4}\left\{\sqrt{1+\sqrt{x^{-4}+1}}-\sqrt{2} \right\} \\=\lim_{x\rightarrow \infty }x^{4}\left\{\sqrt{1+{\frac{x^{-4}}{2}+1}}-\sqrt{2} \right\} \\=\lim_{x\rightarrow \infty }x^{4}\left\{\sqrt{2+{\frac{x^{-4}}{2}}}-\sqrt{2} \right\} \\=\lim_{x\rightarrow \infty }x^{4}\sqrt{2}\left\{\sqrt{1+{\frac{x^{-4}}{4}}}-1 \right\} \\=\lim_{x\rightarrow \infty }x^{4}\sqrt{2}\left\{1+{\frac{x^{-4}}{8}}-1 \right\} \\=\lim_{x\rightarrow \infty }x^{4}\sqrt{2}\left\{{\frac{x^{-4}}{8}} \right\} \\=\frac{\sqrt{2}}{8}=\frac{1}{2\sqrt{2}}