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Question Number 107975 by john santu last updated on 13/Aug/20

$$\frac{\mathrm{♣}JS\mathrm{♡}}{\bullet \equiv \bullet }$$$$\left(1\right)\underset{x\to 0}{\lim }\frac{\sin \left(\tan x\right)-\tan \left(\sin x\right)}{x-\sin x}$$$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }{x}^2\sqrt{(1-\cos \left(\frac{2}{x}\right))\sqrt{(1-\cos \left(\frac{2}{x}\right))\sqrt{(1-\cos \left(\frac{2}{x}\right))\sqrt{...}}}}?$$$$$$

Commented by malwaan last updated on 14/Aug/20

$$fantasticquestionandsolutions$$$$thankyou$$

Answered by bemath last updated on 13/Aug/20

$$\frac{\bullet \mathcal{B}e\mathcal{M}ath\bullet }{joss}$$$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }{x}^2\sqrt{(1-\cos \left(\frac{2}{x}\right))\sqrt{(1-\cos \left(\frac{2}{x}\right))\sqrt{...}}}=L$$$$let\sqrt{(1-\cos \left(\frac{2}{x}\right))\sqrt{(1-\cos \left(\frac{2}{x}\right))\sqrt{...}}}=r$$$$\Rightarrow (1-\cos \left(\frac{2}{x}\right)).r=r^2$$$$\Rightarrow r=1-\cos \left(\frac{2}{x}\right).Then$$$$L=\underset{x\to \mathrm{\infty }}{\lim }{x}^2(1-\cos \left(\frac{2}{x}\right))$$$$let\frac{1}{x}=q\Rightarrow L=\underset{q\to 0}{\lim }\frac{1-\cos 2q}{q^2}$$$$L=\underset{q\to 0}{\lim }\frac{2\sin {}^{2}q}{q^2}=2$$

Answered by bemath last updated on 13/Aug/20

$$\frac{\bullet \mathcal{B}emath\bullet }{⋮⋮}$$$$\left(1\right)\underset{x\to 0}{\lim }\frac{(\tan x-\frac{1}{6}\tan {}^{3}x)-(\sin x+\frac{1}{3}\sin {}^{3}x)}{x-(x-\frac{x^3}{6})}$$$$\underset{x\to 0}{\lim }\frac{(x-\frac{x^3}{6})-(x+\frac{x^3}{3})}{\left(\frac{x^3}{6}\right)}=$$$$\underset{x\to 0}{\lim }\left(\frac{-\frac{3x^3}{6}}{\frac{x^3}{6}}\right)=-3$$

Answered by Dwaipayan Shikari last updated on 13/Aug/20

$$\underset{x\to 0}{\lim }\frac{sin\left(tanx\right)-tan\left(sinx\right)}{x-sinx}$$$$\underset{x\to 0}{\lim }\frac{sinx-tanx}{x-sinx}$$$$\underset{x\to 0}{\lim }\frac{x-\frac{x^3}{6}-x-\frac{x^3}{3}}{x-x+\frac{x^3}{6}}=-3$$$$$$

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