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Question Number 115616 by bobhans last updated on 27/Sep/20

$$\underset{x\to 0}{\lim }\frac{\sqrt[8]{256+\tan x}-\sqrt[3]{1+\sin x}-1}{\sqrt[6]{1+\tan x}-1}?$$

Answered by john santu last updated on 27/Sep/20

$$\underset{x\to 0}{\lim }\frac{2\sqrt[8]{1+\frac{\tan x}{256}}-\sqrt[3]{1+\sin x}-1}{\sqrt[6]{1+\tan x}-1}=$$$$\underset{x\to 0}{\lim }\frac{2(1+\frac{x}{8\times 256})-(1+\frac{x}{3})-1}{(1+\frac{x}{6})-1}=$$$$\underset{x\to 0}{\lim }\frac{x(\frac{1}{4\times 256}-\frac{1}{3})}{\frac{x}{6}}=6.(\frac{1}{4\times 256}-\frac{1}{3})$$$$=\frac{3}{2\times 256}-2=\frac{3}{512}-2$$$$=-\frac{1021}{512}$$

Answered by Dwaipayan Shikari last updated on 27/Sep/20

$$\underset{x\to 0}{\lim }\frac{2\sqrt[8]{1+\frac{\mathrm{tanx}}{256}}-1-\frac{\mathrm{sinx}}{3}-1}{1+\frac{\mathrm{x}}{6}-1}=\frac{\frac{\mathrm{x}}{1024}-\frac{\mathrm{x}}{3}}{\frac{\mathrm{x}}{6}}=\frac{3}{512}-2=\frac{-1021}{512}$$

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