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Question Number 135223 by benjo_mathlover last updated on 11/Mar/21

$$Limit$$$$\left(a\right)\underset{x\to \pi }{\lim }{\tan }^{-1}\left(\tan {}^2\left(\frac{x}{2}\right)\right)=?$$$$\left(b\right)\underset{x\to -1}{\lim }\frac{108(x^2+2x){(x+1)}^3}{{(x^3+1)}^3(x-1)}=?$$

Answered by EDWIN88 last updated on 11/Mar/21

$$\left(\mathrm{b}\right)\underset{x\to -1}{\lim }\frac{108({\mathrm{x}}^2+2\mathrm{x})}{\mathrm{x}-1}\times {\left[\underset{x\to -1}{\lim }\frac{\mathrm{x}+1}{{\mathrm{x}}^3+1}\right]}^3=$$$$\frac{108(-1)}{-2}\times {\left[\underset{x\to -1}{\lim }\frac{\mathrm{x}+1}{(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)}\right]}^3=$$$$54\times {\left[\underset{x\to -1}{\lim }\frac{1}{{\mathrm{x}}^2-\mathrm{x}+1}\right]}^3=54\times \frac{1}{27}=2$$

Answered by liberty last updated on 12/Mar/21

$$\left(a\right)\underset{x\to \pi }{\lim }{\tan }^{-1}\left(\tan {}^2\left(\frac{x}{2}\right)\right)=\underset{x\to \pi }{\lim }{\tan }^{-1}\left(\mathrm{\infty }\right)=\frac{\pi }{2}$$

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