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Question Number 184329 by greougoury555 last updated on 05/Jan/23

$$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}(p+x)-{\tan }^{-1}(p-x)}{x}=?$$

Answered by SEKRET last updated on 05/Jan/23

$$\frac{2}{1+{\mathbf{p}}^2}$$

Answered by mr W last updated on 05/Jan/23

$$y\left(x\right)={\tan }^{-1}x$$$$y^{\prime }\left(x\right)=\frac{1}{1+x^2}$$$$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}(p+x)-{\tan }^{-1}(p-x)}{x}$$$$=\underset{h\to 0}{\lim }\frac{{\tan }^{-1}(p+h)}{h}+\underset{h\to 0}{\lim }\frac{{\tan }^{-1}(p-h)}{-h}$$$$=y^{\prime }\left(x\right){\mid }_{x=p}+y^{\prime }\left(x\right){\mid }_{x=p}$$$$=\frac{1}{1+p^2}+\frac{1}{1+p^2}$$$$=\frac{2}{1+p^2}$$

Answered by cortano1 last updated on 05/Jan/23

$$Find\underset{x\to 0}{\lim }\frac{{\tan }^{-1}(p+x)-{\tan }^{-1}(p-x)}{x}$$$$L=\underset{x\to 0}{\lim }\frac{{\tan }^{-1}(p+x)-{\tan }^{-1}p}{x}+\underset{x\to 0}{\lim }\frac{{\tan }^{-1}p-{\tan }^{-1}(p-x)}{x}$$$$L_1=\underset{x\to 0}{\lim }\frac{{\tan }^{-1}(p+x)-{\tan }^{-1}p}{x}$$$$L_1=\underset{x\to 0}{\lim }\frac{x}{x\{1+p(p+x)\}}=\frac{1}{p^2+1}$$$$L_2=\underset{x\to 0}{\lim }\frac{{\tan }^{-1}p-{\tan }^{-1}(p-x)}{x}$$$$L_2=\underset{x\to 0}{\lim }\frac{x}{x\{1+p(p-x)\}}=\frac{1}{p^2+1}$$$$\therefore L=L_1+L_2=\frac{2}{p^2+1}$$

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