JEE Delight | ISI UGA 2024 | Q23 | "Master Limits with Sandwich Theorem and Definite Integrals

[@harshuldesai8901]

You can form r/n inside the log by adding (r/n)log(n). Split the limit into two: lim_{n->infty} -1/(4log(n)) + lim_{n->infty} (n+2)(n-1)/2n^2 = 0 + 1/2 = 1/2.

[@harshuldesai8901]

The -1/4 comes from the integral

[@Ajay_Vector]

Ya I did the same !

[@rishabhhappy]

@@harshuldesai8901 nahh....... 1/2 hi aata hai from this method also

[@harshuldesai8901]

@@rishabhhappy I meant that the -1/4 in the limit lim_{n->infty} -1/(4log(n)) comes from the integral. The final answer is still 1/2

[@rishabhhappy]

@@harshuldesai8901 achaa bhai sorry i misunderstood.....

[@GovinduNaik-js1gk]

Eagerly waiting for UGB solutions..

[@subhrayanbarman1654]

Another way to solve this:
Let, An= 2log2+3log3+...+nlogn
and, Bn= n²logn
Then, ∆A=An-An-1=nlogn
∆B=Bn-Bn-1=n²logn-(n-1)²log(n-1) ~(2n-1)logn
∆A/∆B =1/2

[@mystik4957]

thats so good bruh wtf. This is basically kind of like l'hospital for discrete sets of data

[@KALAMKAAR_Official]

But in this solution how can you say lim n-- inf n² ln(1+1/n) is zero ?
Edit got it ln(1+x) = x se ye n likhha fir Inf/inf ho rha tha to continuous L hopital use kar liya

[@harshuldesai8901]

@@mystik4957 Stolz-Cesaro theorem. It's a test for series convergence and it's basically L'H for series. Basically says that lim_{n->infty} (a_n - a_{n-1})/(b_n - b_{n-1}) = lim_{n->infty} a_n/b_n where b_n is a strictly increasing function of n.

[@shivanshnigam4015]

Sir I turned the numerator in to a definite integral but with some ln(n) terms remaining so just wrote the definite integral as some constant then used LHopital and got 1/2 as the ans

[@piyushraj760]

stolz cesaro is also a good approach sir

[@divyakumar8147]

thanks sir

[@edu_in_iitg]

Handwriting dekhar ulti aa gayi re bhaiya 😅 anyways gr8 soln

[@Your_Study_Buddy_SD]

This question seemed to be easy at first look but after a couple of calculations , realization strikes that oh boy this will take a twist.
This is a wonderful question.🤌👌