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Question Number 199451 by hardmath last updated on 03/Nov/23

$$\mathrm{Find}:$$$$1.\underset{\mathbf{n}\to \mathrm{\infty }}{\lim }\sqrt[5\mathbf{n}]{\frac{5\mathrm{n}-25}{3\mathrm{n}+15}}$$$$2.\underset{\mathbf{x}\to \mathrm{\infty }}{\lim }(\sqrt[3]{{\mathrm{x}}^3+{3\mathrm{x}}^2+1}-\sqrt[3]{{\mathrm{x}}^3-{3\mathrm{x}}^2+1})$$$$3.\underset{\mathbf{x}\to 0}{\lim }\frac{\cos {4\mathrm{x}}^3-1}{{\sin }^{6}2\mathrm{x}}$$

Answered by cortano12 last updated on 04/Nov/23

$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }\frac{{6\mathrm{x}}^2}{\sqrt[3]{{({\mathrm{x}}^3+{3\mathrm{x}}^2+1)}^2}+\sqrt[3]{({\mathrm{x}}^3+{3\mathrm{x}}^2+1)({\mathrm{x}}^3-{3\mathrm{x}}^2+1)}+\sqrt[3]{{({\mathrm{x}}^3-{3\mathrm{x}}^2+1)}^2}}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{{6\mathrm{x}}^2}{{\mathrm{x}}^2[\sqrt[3]{{(1+\frac{3}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^3})}^2}+\sqrt[3]{(1-\frac{3}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^3})(1+\frac{3}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^3})}+\sqrt[3]{{(1-\frac{3}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^3})}^2}]}$$$$=\frac{6}{3}=2$$$$$$

Commented by hardmath last updated on 04/Nov/23

$$thankyouser$$

Answered by cortano12 last updated on 04/Nov/23

$$\left(3\right)\underset{x\to 0}{\lim }\frac{\cos {4\mathrm{x}}^3-1}{\sin {}^{6}2\mathrm{x}}$$$$=\underset{x\to 0}{\lim }\frac{-\sin {}^2\left({4\mathrm{x}}^3\right)}{(\cos {4\mathrm{x}}^3+1)\sin {}^6\left(2\mathrm{x}\right)}$$$$=-\frac{1}{2}\underset{x\to 0}{\lim }\frac{{\left({4\mathrm{x}}^3\right)}^2}{{\left(2\mathrm{x}\right)}^6}=-\frac{1}{2}.\frac{16}{64}$$$$=-\frac{1}{8}$$$$$$

Commented by hardmath last updated on 04/Nov/23

$$thankyouser$$

Answered by cortano12 last updated on 04/Nov/23

$$\left(1\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{5\mathrm{n}-25}{3\mathrm{n}+15}\right)}^{\frac{1}{5\mathrm{n}}}$$$$=\mathrm{e}{}^{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\frac{5\mathrm{n}-25}{3\mathrm{n}+15}-1).\frac{1}{5\mathrm{n}}}$$$$=\mathrm{e}{}^{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left(\frac{2\mathrm{n}-40}{15\mathrm{n}(\mathrm{n}+5)}\right)}={\mathrm{e}}^0=1$$

Commented by hardmath last updated on 04/Nov/23

$$thankyouser$$

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