granice
: 15 cze 2018, 16:42
Obliczyć granice \(\lim _{x\to \infty }\left(x^3 \left(\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2}x \right) \right)\)
\(=\lim _{x\to \infty }\left(x^3 \left(\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2}x \right) \right) \cdot \frac{\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2}x }{\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2}x } =\lim _{x\to \infty }x^3 \cdot \frac{\sqrt{x^4+1}-x^2}{x(\sqrt{1+\sqrt{1+ \frac{1}{x^4} }}+\sqrt{2} )} =\\=
\lim _{x\to \infty }x^2 \cdot \frac{(\sqrt{x^4+1}-x^2) \cdot \frac{\sqrt{x^4+1}+x^2}{\sqrt{x^4+1}+x^2} }{\sqrt{1+\sqrt{1+ \frac{1}{x^4} }}+\sqrt{2} }=
\lim _{x\to \infty }x^2 \cdot \frac{ \frac{1}{x^2(\sqrt{1+ \frac{1}{x^4} }+1)} }{\sqrt{1+\sqrt{1+ \frac{1}{x^4} }}+\sqrt{2} }= \frac{ \frac{1}{2} }{ \sqrt{2}+ \sqrt{2} } = \frac{ \sqrt{2} }{8}\)
