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Question Number 133114 by bemath last updated on 19/Feb/21

$$\underset{x\to \mathrm{\infty }}{\lim }{(\frac{\sqrt{{\mathrm{x}}^2-\mathrm{x}}}{\mathrm{x}}+\frac{1}{4}\sin \left(\frac{2}{\mathrm{x}}\right))}^{{\mathrm{x}}^2+\sin 3\mathrm{x}}?$$

Answered by bobhans last updated on 19/Feb/21

$$L=\underset{x\to \mathrm{\infty }}{\lim }{(\frac{\sqrt{x^2-x}}{x}+\frac{1}{4}\sin \left(\frac{2}{x}\right))}^{x^2+\sin \left(3x\right)}$$$$=\underset{x\to \mathrm{\infty }}{\lim }{(\sqrt{1-\frac{1}{x}}+\frac{1}{4}\sin \left(\frac{2}{x}\right))}^{x^2+\sin \left(3x\right)}$$$$\mathrm{let}\frac{1}{x}=z,z\to 0^{+}$$$$=\underset{z\to 0^{+}}{\lim }{(\sqrt{1-z}+\frac{1}{4}\sin \left(2z\right))}^{\frac{1}{z^2}+\sin \left(\frac{3}{z}\right)}$$$$\ln L=\underset{z\to 0^{+}}{\lim }(\frac{1}{z^2}+\sin \left(\frac{3}{z}\right))\ln (\sqrt{1-z}+\frac{1}{4}\sin \left(2z\right))$$$$=\underset{z\to 0^{+}}{\lim }(\frac{(1+z^2\sin \left(\frac{3}{z}\right))\ln (\sqrt{1-z}+\frac{1}{4}\sin 2z)}{z^2}$$$$bySandwichtheoremwegetthat$$$$\underset{z\to 0^{+}}{\lim }(1+z^2\sin \left(\frac{3}{z}\right))=1$$$$thereforethelimitreducesto$$$$\ln L=\underset{z\to 0^{+}}{\lim }\frac{\ln (\sqrt{1-z}+\frac{1}{4}\sin \left(2z\right))}{z^2}$$$$\ln L=\underset{z\to 0^{+}}{\lim }\frac{-\frac{1}{2}{(1-z)}^{-\frac{1}{2}}+\frac{1}{2}\cos \left(2t\right)}{2t(\sqrt{1-z}+\frac{1}{4}\sin 2z)}$$$$=\underset{z\to 0^{+}}{\lim }\frac{1}{\sqrt{1-z}+\frac{1}{4}\sin 2z}.\underset{z\to 0^{+}}{\lim }\frac{-{(1-z)}^{-1/2}+\cos 2z}{4z}$$$$=1.\underset{z\to 0^{+}}{\lim }\frac{-\frac{1}{2}{(1-z)}^{-3/2}-2\sin 2z}{4}$$$$=-\frac{1}{8};\ln L=-\frac{1}{8}thenL=e^{-1/8}$$$$orL=\frac{1}{\sqrt[8]{e}}$$

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