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Question Number 128908 by bramlexs22 last updated on 11/Jan/21

$$\underset{x\to \mathrm{\infty }}{\lim }(\mathrm{x}\sqrt{{\mathrm{x}}^2+4}-\sqrt{{\mathrm{x}}^4+16})=?$$

Answered by Dwaipayan Shikari last updated on 11/Jan/21

$$(x^2\sqrt{1+\frac{4}{x^2}}-x^2\sqrt{1+\frac{16}{x^4}})=x^2(1+\frac{2}{x^2}-1-\frac{8}{x^4})=2$$

Answered by liberty last updated on 11/Jan/21

$$\mathrm{standard}\mathrm{way}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{(\sqrt{{\mathrm{x}}^4+{4\mathrm{x}}^2}-\sqrt{{\mathrm{x}}^4+16})(\sqrt{{\mathrm{x}}^4+{4\mathrm{x}}^2}+\sqrt{{\mathrm{x}}^4+16})}{\sqrt{{\mathrm{x}}^4+{4\mathrm{x}}^2}+\sqrt{{\mathrm{x}}^4+16}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{4\mathrm{x}}^2-16}{{\mathrm{x}}^2(\sqrt{1+{4\mathrm{x}}^{-2}}+\sqrt{1+{16\mathrm{x}}^{-4}})}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{4-{16\mathrm{x}}^{-2}}{\sqrt{1+{4\mathrm{x}}^{-2}}+\sqrt{1+{16\mathrm{x}}^{-4}}}=\frac{4}{2}=2$$

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