Math, asked by Anonymous, 8 months ago

EVALUATE.............

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Answered by Anonymous
6

 \huge\bf\red{AnswEr:-}

limp~∞ [(2x-1)/2x+1]^2x

limp~∞ [(2-1/x)/(2+1/x)]^2x [taking x common]

Putting limit shows 1^∞ form

For limp~∞ f(x)^g(x) 1^∞ form we can directly write

limp~∞ e^g(x)[f(x) - 1]

So limp~∞ [(2x-1)/2x+1]^2x

or limp~∞ e^2x[(2x-1)/(2x+1) - 1]

or limp~∞ e^2x[(2x-1-2x-1)/(2x+1)]

or limp~∞ e^(-4x/2x+1)

or limp~∞ e^[-4/(2+1/x)] (taking x common)

Putting the limit x~∞, 1/x ~0

or limp~∞ e^[-4/2] =e^-2

\bigstarSo answer =e^-2\bigstar

_______\bigstar________

Answered by HariesRam
58
limp~∞ [(2x-1)/2x+1]^2x

limp~∞ [(2-1/x)/(2+1/x)]^2x [taking x common]

Putting limit shows 1^∞ form

For limp~∞ f(x)^g(x) 1^∞ form we can directly write

limp~∞ e^g(x)[f(x) - 1]

So limp~∞ [(2x-1)/2x+1]^2x

or limp~∞ e^2x[(2x-1)/(2x+1) - 1]

or limp~∞ e^2x[(2x-1-2x-1)/(2x+1)]

or limp~∞ e^(-4x/2x+1)

or limp~∞ e^[-4/(2+1/x)] (taking x common)

Putting the limit x~∞, 1/x ~0

or limp~∞ e^[-4/2] =e^-2


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