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Question Number 148819 by mathlove last updated on 31/Jul/21

Answered by liberty last updated on 31/Jul/21

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}}{(1+\frac{{\mathrm{x}}^2}{\mathrm{n}})-(1+\frac{\mathrm{x}}{\mathrm{n}})}=$$$$\underset{x\to 0}{\lim }\frac{\mathrm{x}}{\frac{\mathrm{x}}{\mathrm{n}}(\mathrm{x}-1)}=\underset{x\to 0}{\lim }\frac{\mathrm{n}}{\mathrm{x}-1}=-\mathrm{n}$$

Answered by Kamel last updated on 31/Jul/21

$$L=\underset{x\to 0}{lim}\frac{1}{\frac{\sqrt[n]{1+x^2}-1}{x}-\frac{\sqrt[n]{1+x}-1}{x}}=\frac{1}{-\frac{1}{n}}=-n$$

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