\lim_(x->\infty )(1+(1)/(2x-2))^(x-1) without using L hospital rule
\lim_(x->\infty )(1+(1)/(2x-2))^(x-1) without using L hospital rule
Narayana Murthy Pola , 6 Years ago
Grade 11
Differential Calculus> \lim_(x->\infty )(1+(1)/(2x-2))^(x-1) wit...
4 Answers
Arun
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Vikas TU
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Aditya Gupta
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 get
L= limit (y->infty )(1+1/2y)^y
logL= limit (y->infty )y*log(1+1/2y)
put 1/2y= z
so logL= limit (z ->0)log(1+z)/2z
or 2logL= limit (z ->0)log(1+z)/z= 1
or logL= ½
or L= e^(1/2)= sqrt(e)
KINDLY APPROVE :))
Manish kumar
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thanks
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