This qsn was asked in 1 of our school exams......
Find limit as n→∞.....
\frac{\sqrt{(n+1)}+\sqrt{(n+2)}+....\sqrt{2n}}{n\sqrt{n}}
Applicaion of integral calculus gives ans as
\frac{2}{3}(2\sqrt{2}+1)
While if we solve by the normal differential cal. way, ans turns out to be 0......Which is correct and why is the other wrong?
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8 Answers
66The correct answer is \dfrac{2}{3}(2\sqrt{2}-1) which is \int_0^1 \sqrt{1+x}\ \mathrm dx which is what the given limit is.
And can you tell how are you getting 0 by the "differential calculus way".
62are you guys trying to differentiate?? wrt n?
If that is the case then you cant .. bcos this is not a differentiable function(since it is not a continuous function!)
11Yes there's a minus in place of that plus by integration....sorry....
What I did was to take the general term as \sqrt{\frac{n+k}{n^3}}=\sqrt{\frac{1}{n^2}+\frac{k}{n^3}}=0 as n →∞.....
Here k varies from 1 to n.....and thus I got it to be 0....
1but if k varies from 0 to n then how did u get zero ...
and what has this to do with diff calculus ?
62but there are n of them....
And if you infinitely many (n) terms which are close to zero.. the answer may not be zero.. :)