Question

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

Answer 1

$\huge\bf\red{AnswEr:-}$

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

$\bigstar$So answer =e^-2$\bigstar$

_______$\bigstar$________

Answer 2

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