JEE Main 2021 (Online) 27th July Morning ShiftQuestion 1
The value of $\lim_{n\to\infty}\frac1n\sum_{j=1}^{n}\frac{(2j-1)+8n}{(2j-1)+4n}$ is equal to :
Definite Integration
Solution
Concept. The points $t_j=\frac{2j-1}n$ are midpoints of $n$ equal parts of $[0,2]$, each of width $\frac2n$. Formulas used. (1) $\lim\frac2n\sum_{j=1}^ng(t_j)=\int_0^2g(t)\,dt$Step 1 - Rewrite. Divide top and bottom by $n$: $$\frac1n\sum_{j=1}^n\frac{t_j+8}{t_j+4}=\frac12\cdot\frac2n\sum_{j=1}^n\frac{t_j+8}{t_j+4}.$$ By (1), the limit is $\frac12\int_0^2\frac{t+8}{t+4}dt$. Step 2 - Integrate. $\frac{t+8}{t+4}=1+\frac4{t+4}$: $$\frac12\Big[t+4\ln(t+4)\Big]_0^2=\frac12\Big(2+4\ln\frac64\Big)=1+2\log_e\frac32.$$ Answer: (B)