The value of $\lim_{n\to\infty}\dfrac{\left(\lim_{x\to 0^-}\sum_{r=1}^{2n+1}[x^r]\right)+n+3}{\ln(1+n)^n-\ln(n)^n}$ (where $[.]$ denotes GIF)
Step-by-Step Solution
Key Concept: For $x\to 0^-$: $[x^r]=-1$ for odd $r$, $[x^r]=0$ for even $r$
$\sum[x^r]$ for $x\to 0^-$: odd powers $r$: $[x^r]=-1$ ($n+1$ such terms); even: $0$. Sum $=-(n+1)$. Numerator: $-(n+1)+n+3=2$. Denominator: $\ln(1+n)^n-n\ln n=n\ln\frac{n+1}{n}\to 1$. Limit $=2$.
Correct Answer: 2