The value of the integer $k$ such that the following limit $L$ is finite and non-zero:
$$L = \lim_{x\to 0}\frac{\ln(x^x)\cdot(e^{x^x-1}-1)-(x^x-1)}{\sin^k(x\ln x)}$$
is:
Step-by-Step Solution
Key Concept: Let $t = x\ln x$, so $x^x = e^t$ and as $x\to 0^+$, $t\to 0$. Then $\ln(x^x)=t$, $x^x-1=e^t-1\approx t$, $e^{x^x-1}-1=e^{e^t-1}-1\approx e^t-1$. Numerator $\approx t(e^t-1)-(e^t-1) = (t-1)(e^t-1)$... leading behavior is $-t$ as $t\to 0$ (since $e^t-1\approx t$ and $(t-1)t\approx -t$).
Let $t=x\ln x\to 0$. $N = t(e^{e^t-1}-1)-(e^t-1)\approx t\cdot t-t = t^2-t\sim -t$ as $t\to 0$. Denominator $\sim t^k$. For finite nonzero: $k=1$, $L=-1$. Answer: $k=\mathbf{1}$.
Correct Answer: 1