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Question Number 88580 by jagoll last updated on 11/Apr/20

$$$$$$\underset{x\to 0}{\lim }\frac{\sqrt{1+4\mathrm{x}}-\sqrt[3]{1+6\mathrm{x}}}{1-\cos 3\mathrm{x}}=$$

Commented by mathmax by abdo last updated on 11/Apr/20

$$letf\left(x\right)=\frac{\sqrt{1+4x}{-}^3\sqrt{1+6x}}{1-cos\left(3x\right)}wehave{(1+4x)}^{\frac{1}{2}}\sim 1+\frac{1}{2}\left(4x\right)$$$$+\frac{1}{2}\left(\frac{1}{2}\right)(\frac{1}{2}-1){\left(4x\right)}^2=1+2x-\frac{1}{8}\left(16x^2\right)=1+2x-2x^2$$$${(1+6x)}^{\frac{1}{3}}=1+\frac{1}{3}\left(6x\right)+\frac{1}{2}\left(\frac{1}{3}\right)(\frac{1}{3}-1){\left(6x\right)}^2$$$$=1+2x-\frac{2}{18}\left(36\right)x^2=1+2x-4x^2$$$$1-cos\left(3x\right)\sim \frac{9x^2}{2}\Rightarrow f\left(x\right)\sim \frac{1+2x-2x^2-1-2x+4x^2}{\frac{9x^2}{2}}=\frac{4x^2}{9x^2}$$$$\Rightarrow {lim}_{x\to 0}f\left(x\right)=\frac{4}{9}$$$$$$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{i}\mathrm{got}0.\mathrm{it}\mathrm{correct}?$$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{how}\mathrm{get}\sqrt{1+4\mathrm{x}}=1+\frac{1}{2}\left(4\mathrm{x}\right)+\frac{1}{2}\times \frac{1}{2}(\frac{1}{2}-1){\left(4\mathrm{x}\right)}^2$$$$\mathrm{Maclaurin}\mathrm{series}?$$

Commented by mathmax by abdo last updated on 11/Apr/20

$$yes{(1+u)}^{\alpha }=1+\frac{\alpha }{1!}u+\frac{\alpha (\alpha -1)}{2!}{u}^2+...andtakeu=4xand\alpha =\frac{1}{2}$$

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