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Question Number 161770 by cortano last updated on 22/Dec/21

$$\underset{x\to 0}{\lim }\frac{{(\sqrt{1+2x^2}+2x)}^{2021}-{(\sqrt{1+2x^2}-2x)}^{2021}}{x}$$

Answered by Ar Brandon last updated on 22/Dec/21

$$$$$$a^n-b^n=(a-b)\underset{k=0}{\sum ^{n-1}}{a}^{n-1-k}{b}^k$$$$\mathcal{L}=\underset{x\to 0}{\lim }\frac{{(\sqrt{1+2x^2}+2x)}^{2021}-{(\sqrt{1+2x^2}-2x)}^{2021}}{x}$$$$=\underset{x\to 0}{\lim }\frac{4x\underset{k=0}{\sum ^{2020}}{(\sqrt{1+2x^2}+2x)}^{2020-k}{(\sqrt{1+2x^2}-2x)}^k}{x}$$$$=4\underset{x\to 0}{\lim }\underset{k=0}{\sum ^{2020}}{(\sqrt{1+2x^2}+2x)}^{2020-k}{(\sqrt{1+2x^2}-2x)}^k$$$$=4\underset{k=0}{\sum ^{2020}}\left(1\right)=4\left(2021\right)=8084$$

Answered by qaz last updated on 22/Dec/21

$$\underset{\mathrm{x}\to 0}{\lim }\frac{{(\sqrt{1+2x^2}+2x)}^{2021}-{(\sqrt{1+2x^2}-2x)}^{2021}}{x}$$$$=\underset{\mathrm{x}\to 0,\xi \to 1}{\lim }\frac{2021{\xi }^{2020}}{\mathrm{x}}\left(4\mathrm{x}\right)$$$$=8084$$

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