lim_(x→a) (2−(x/a))^(tan (((πx)/(2a)))) =?


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Question Number 115541 by bobhans last updated on 26/Sep/20

$\lim_{x \to a} {(2 - \frac{x}{a})}^{\tan (\frac{π x}{2 a})} = ?$
	Answered by bemath last updated on 26/Sep/20

	$L = \lim_{x \to a} {(2 - \frac{x}{a})}^{\tan (\frac{π x}{2 a})}$ $\ln L = \lim_{x \to a} \frac{\ln (2 - \frac{x}{a})}{\cot (\frac{π x}{2 a})}$ $\ln L = \lim_{x \to a} \frac{- \frac{1}{a}}{2 - \frac{x}{a}} . \frac{1}{(- \frac{π}{2 a} \csc^{2} (\frac{π x}{a}))}$ $\ln L = \lim_{x \to a} \frac{2 . \sin^{2} (\frac{π x}{2 a})}{π (\frac{2 a - x}{a})} = \frac{2}{π}$ $L = e^{\frac{2}{π}} = \sqrt[π]{e^{2}}$
	Answered by Dwaipayan Shikari last updated on 26/Sep/20

	$\lim_{x \to a} {(1 + 1 - \frac{x}{a})}^{\tan (\frac{π x}{2 a})} = y$ $\underset{x \to a}{\lim t a n} (\frac{π x}{2 a}) \log (1 + 1 - \frac{x}{a}) = logy$ $(1 - \frac{x}{a}) \tan (\frac{π x}{2 a}) = logy$ $(1 - \frac{x}{a}) \sin (\frac{π x}{2 a}) . \frac{1}{\cos (\frac{π x}{2 a})} = logy$ $(1 - \frac{x}{a}) \frac{1}{\sin (\frac{π}{2} - \frac{π x}{2 a})} = logy$ $\frac{(1 - \frac{x}{a})}{\frac{π}{2} (1 - \frac{x}{a})} = logy$ $y = e^{\frac{2}{π}}$
	Answered by mnjuly1970 last updated on 26/Sep/20
	$solution: 1−(x/a) =t {_(t→0) ^(x→a) lim_(t→0) (1+t)^(tan[(π/2)(1−t)]) =lim_(t→0 ) (1+t)^(cot((π/2)t)) =_(t→0) ^( sin((π/2)t)∽(π/2)t) lim_(t→0) [((1+t)^(2/π) )^(1/t) ]^(cos((π/2)t)) =_(t→0) ^(cos((π/2)t)→1) =e^(2/π) ✓ m.n.july.1970$
	$s o l u t i o n :$ $1 - \frac{x}{a} = t {_{t \to 0}^{x \to a}$ ${l i m}_{t \to 0} {(1 + t)}^{t a n [\frac{π}{2} (1 - t)]}$ $= {l i m}_{t \to 0} {(1 + t)}^{c o t (\frac{π}{2} t)} \underset{t \to 0}{\overset{s i n (\frac{π}{2} t) ∽ \frac{π}{2} t}{=}}$ ${l i m}_{t \to 0} {[{({(1 + t)}^{\frac{2}{π}})}^{\frac{1}{t}}]}^{c o s (\frac{π}{2} t)} \underset{t \to 0}{\overset{c o s (\frac{π}{2} t) \to 1}{=}}$ $= e^{\frac{2}{π}} ✓$ $m . n . j u l y . 1970$
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