lim_(x→(π/(2n))) ((√((sin 2nx)/(1+cos nx)))/(4n^2 x^2 −π^2 )) =?


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Question Number 198572 by cortano12 last updated on 22/Oct/23

$\lim_{x \to \frac{π}{2 n}} \frac{\sqrt{\frac{\sin 2 nx}{1 + \cos nx}}}{{4 n}^{2} x^{2} - π^{2}} = ?$
	Answered by MM42 last updated on 22/Oct/23
	$(π/(2n))−x=u lim_(u→0) ((√((sin2n((π/(2n))−u))/(1+cosn((π/(2n))−u))))/((2n((π/(2n))−u)−π)(2n((π/(2n))−u)+π))) =lim_(u→0) ((√((sin(2nu))/(1+sin(nu))))/((−2nu)(2π−2nu))) =lim_(u→0) ((√((sin(2nu))/(1+sin(nu))))/((−2nu)(2π−2nu)))= { ((+∞ if u→0^− )),((−∞ if u→0^+ )) :}$
	$\frac{π}{2 n} - x = u$ ${l i m}_{u \to 0} \frac{\sqrt{\frac{s i n 2 n (\frac{π}{2 n} - u)}{1 + c o s n (\frac{π}{2 n} - u)}}}{(2 n (\frac{π}{2 n} - u) - π) (2 n (\frac{π}{2 n} - u) + π)}$ $= {l i m}_{u \to 0} \frac{\sqrt{\frac{s i n (2 n u)}{1 + s i n (n u)}}}{(- 2 n u) (2 π - 2 n u)}$ $= {l i m}_{u \to 0} \frac{\sqrt{\frac{s i n (2 n u)}{1 + s i n (n u)}}}{(- 2 n u) (2 π - 2 n u)} = {\begin{cases} + \infty i f u \to 0^{-} \\ - \infty i f u \to 0^{+} \end{cases}$
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