JEE Main 2023 — Limits Question with Solution

The given limit $\underset{n\to \infty }{\lim }\frac{1+2-3+4+5-6+\dots +(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3-}\sqrt{n^4+5n+4}}$ can be expressed as,$\underset{n\to \infty }{\lim }\frac{{\displaystyle \sum _{r=1}^n}\left(3r-2\right)+\left(3r-1\right)-3r}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$

$=\underset{n\to \infty }{\lim }\frac{{\displaystyle \sum _{r=1}^{n}3\left(r-1\right)}}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$

$=\underset{n\to \infty }{\lim }\frac{0+3+6+9+\dots .n terms}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}.......\left(1\right)$

We know the sum of the nth term of A.P. is given by

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Using above formula in equation(1) we get,

$=\underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{3n(n-1)}{2}}}{\left(\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}\right)}$

$=\underset{n\to \infty }{\lim }\frac{{\displaystyle 3n(n-1)}}{2\left(\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}\right)}$

$=\underset{n\to \infty }{\lim }\frac{{\displaystyle 3(1-\frac{1}{n})}}{2\left(\sqrt{2+{\displaystyle \frac{4}{n^3}}+{\displaystyle \frac{3}{n^4}}}-\sqrt{1+{\displaystyle \frac{5}{n^3}}+{\displaystyle \frac{4}{n^4}}}\right)}$

$=\frac{{\displaystyle 3(1-0)}}{2\left(\sqrt{2+{\displaystyle 0}+{\displaystyle 0}}-\sqrt{1+{\displaystyle 0}+0}\right)}$

$=\frac{3}{2(\sqrt{2}-1)}=\frac{3}{2}(\sqrt{2}+1)$