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Question Number 213073 by ajfour last updated on 29/Oct/24

$$L=\underset{x\to 0}{\lim }\left\{\frac{1}{x^2}{\tan }^{-1}(\sqrt{1+\frac{x^2}{a^2}}-1)\right\}$$

Answered by universe last updated on 29/Oct/24

$$L=\underset{x\to 0}{\lim }\frac{\sqrt{1+\frac{x^2}{a^2}}-1}{x^2}$$$$L=\frac{1}{a}\underset{x\to 0}{\lim }\frac{\sqrt{a^2+x^2}-a}{x^2}$$$$L=\frac{1}{a}\underset{x\to 0}{\lim }\frac{2x/2\sqrt{a^2+x^2}}{2x}$$$$L=1/2a^2$$$$$$

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