Please solve the 25th question as early as possible.

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Aditya Gupta · Answer

hello samruddhi. note that we can write f(x) as: f(x)= Lt n tends to infinity [1/{1+(lnx)^n}] now, e obvious lies in between [1, 3] as its approx value is 2.71828..... so, when x belongs to [1, e), lnx shall lie in [0, 1), which means that lnx is less than 1. and anything less than 1 raised to power n when n tends to infinity is 0. so, f(x)= 1/(1+0) or f(x)= 1 for x in [1, e] f(x)= 1/(1+1)= ½ for x= e (since lnx= 1 when x=e) and when x is greater than e. lnx will be greater than lne= 1. since anything greater than 1 raised to power n when n tends to infinity is infinity. so, f(x)= 1/(1+inf) or f(x)= 0 for x in (e, 3]. clearly f is not diff at x= e (=K) so, [K^2]= [e^2]= [2.71828...^2] = [7.38905....] = 7 kindly approve :))

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Please solve the 25th question as early as possible.

Please solve the 25th question as early as possible.

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