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Question Number 46136 by Meritguide1234 last updated on 21/Oct/18

Answered by ajfour last updated on 21/Oct/18

$${\left(\cos nx\right)}^{1/n}={(1-2\sin {}^2\frac{nx}{2})}^{1/n}$$$$if\underset{x\to 0}{\lim }then=1-\frac{2}{n}\left(\frac{n^2{x}^2}{4}\right)$$$$=1-\frac{{nx}^2}{2}$$$$L=\frac{1}{2}[\frac{n(n+1)}{2}-1]$$$$L=\frac{n^2+n-2}{4}$$

Commented by Meritguide1234 last updated on 21/Oct/18

$$nice$$

Commented by Meritguide1234 last updated on 21/Oct/18

Commented by rahul 19 last updated on 21/Oct/18

$$Ajfoursir,plsexplainthislastline.$$$$HowyougotL=\frac{1}{2}[n(n+1)/2-1]..??$$

Commented by Meritguide1234 last updated on 21/Oct/18

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