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Question Number 199393 by Calculusboy last updated on 02/Nov/23

Answered by cortano12 last updated on 03/Nov/23

$$\mathrm{A}=\underset{x\to 0}{\lim }\frac{(1+\frac{\mathrm{x}}{2011})(1+\frac{2\mathrm{x}}{2012})(1+\frac{3\mathrm{x}}{2013})-1}{\mathrm{x}}$$$$\mathrm{A}=\frac{1}{2011}+\frac{2}{2012}+\frac{3}{2013}$$

Commented by essaad last updated on 03/Nov/23

hi sir Can you explain to us how you did that?

Commented by MathematicalUser2357 last updated on 03/Nov/23

$$$$Calculator???

Answered by mr W last updated on 03/Nov/23

$$\sqrt[2011]{1+x}={(1+x)}^{\frac{1}{2011}}=1+\frac{x}{2011}+termswith{x}^{2}orhigher$$$$=1+\frac{x}{2011}+o\left(x^2\right)$$$$similarly$$$$\sqrt[2012]{1+2x}=1+\frac{2x}{2012}+o\left(x^2\right)$$$$\sqrt[2013]{1+3x}=1+\frac{3x}{2013}+o\left(x^2\right)$$$$$$$$\sqrt[2011]{1+x}\times \sqrt[2012]{1+2x}\times \sqrt[2013]{1+3x}$$$$=[1+\frac{x}{2011}+o\left(x^2\right)][1+\frac{2x}{2012}+o\left(x^2\right)][1+\frac{3x}{2013}+o\left(x^2\right)]$$$$=1+(\frac{1}{2011}+\frac{2}{2012}+\frac{3}{2013})x+o\left(x^2\right)$$$$$$$$\frac{\sqrt[2011]{1+x}\times \sqrt[2012]{1+2x}\times \sqrt[2013]{1+3x}-1}{x}$$$$=(\frac{1}{2011}+\frac{2}{2012}+\frac{3}{2013})+o\left(x\right)$$$$$$$$\underset{x\to 0}{\lim }\frac{\sqrt[2011]{1+x}\times \sqrt[2012]{1+2x}\times \sqrt[2013]{1+3x}-1}{x}$$$$=\underset{x\to 0}{\lim }[(\frac{1}{2011}+\frac{2}{2012}+\frac{3}{2013})+o\left(x\right)]$$$$=\frac{1}{2011}+\frac{2}{2012}+\frac{3}{2013}+\underset{x\to 0}{\lim }o\left(x\right)$$$$=\frac{1}{2011}+\frac{2}{2012}+\frac{3}{2013}+0$$$$=\frac{1}{2011}+\frac{2}{2012}+\frac{3}{2013}$$

Commented by Calculusboy last updated on 03/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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