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Question Number 131370 by EDWIN88 last updated on 10/Feb/21

$$\underset{x\to -\mathrm{\infty }}{\lim }\sqrt{4x^4+6x^2}+x\sqrt{4x^2+2}=?$$

Answered by JDamian last updated on 04/Feb/21

$$\mathrm{\infty }$$

Answered by EDWIN88 last updated on 10/Feb/21

$$\underset{x\to -\mathrm{\infty }}{\lim }\sqrt{{\mathrm{x}}^2({4\mathrm{x}}^2+6)}+\mathrm{x}\sqrt{{\mathrm{x}}^2(4+\frac{2}{{\mathrm{x}}^2})}=$$$$\underset{x\to -\mathrm{\infty }}{\lim }-\mathrm{x}\sqrt{{4\mathrm{x}}^2+6}-{\mathrm{x}}^2\sqrt{4+\frac{2}{{\mathrm{x}}^2}}=$$$${\underset{x\to -\mathrm{\infty }}{\lim x}}^2\sqrt{4+\frac{6}{{\mathrm{x}}^2}}-{\mathrm{x}}^2\sqrt{4+\frac{2}{{\mathrm{x}}^2}}=$$$$\mathrm{let}\frac{1}{{\mathrm{x}}^2}=\mathrm{h}\wedge \mathrm{h}\to 0$$$$\underset{\mathrm{h}\to 0}{\lim }\frac{\sqrt{4+{6\mathrm{h}}^2}-\sqrt{4+{2\mathrm{h}}^2}}{{\mathrm{h}}^2}=2\times \underset{\mathrm{h}\to 0}{\lim }\frac{\sqrt{1+\frac{3}{2}{\mathrm{h}}^2}-\sqrt{1+\frac{{\mathrm{h}}^2}{2}}}{\mathrm{h}}$$$$=2\times \underset{\mathrm{h}\to 0}{\lim }\frac{(1+\frac{{3\mathrm{h}}^2}{4})-(1+\frac{{\mathrm{h}}^2}{4})}{{\mathrm{h}}^2}=1$$

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