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`        Please solve the 25th question as early as possible.`
one month ago

```							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....]= 7kindly approve :))
```
one month ago
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