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\lim_(x->\infty )(1+(1)/(2x-2))^(x-1) without using L hospital rule
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your ques is to find limit (x->infty )(1+(1)/(2x-2))^(x-1)= limit (x->infty )(1+1/2(x-1))^(x-1)let y= x – 1, we getL= limit (y->infty )(1+1/2y)^ylogL= limit (y->infty )y*log(1+1/2y)put 1/2y= zso logL= limit (z ->0)log(1+z)/2zor 2logL= limit (z ->0)log(1+z)/z= 1or logL= ½ or L= e^(1/2)= sqrt(e)KINDLY APPROVE :))
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