EVALUATE.............
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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
So answer =e^-2
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Answered by
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
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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