JEE Main 2023 (Online) 25th January Morning ShiftQuestion 1
The value of $\lim_{n\to\infty}\frac{1+2-3+4+5-6+\ldots+(3n-2)+(3n-1)-3n}{\sqrt{2n^{4}+4n+3}-\sqrt{n^{4}+5n+4}}$ is
Limits & Continuity
Solution
Each block gives $(3k-2)+(3k-1)-3k=3k-3$, so the numerator is $\frac{3n(n-1)}2$. The denominator $\sim(\sqrt2-1)n^2$, so the limit is $\frac{3}{2(\sqrt2-1)}=\frac32(\sqrt2+1)$.