\(\displaystyle\lim_{x\to 0}\left(1^{\csc^2 x}+2^{\csc^2 x}+3^{\csc^2 x}+\ldots+100^{\csc^2 x}\right)^{\sin^2 x}\)
100
Let \(k=\csc^2 x\); as \(x\to 0\), \(k\to \infty\) and \(\sin^2x=1/k\).
In the sum \(1^{k}+2^{k}+\ldots+100^{k}\), the term \(100^{k}\) dominates all others as \(k\to\infty\), since \((m/100)^{k}\to 0\) for every \(m<100\).
So the sum can be written as \(100^{k}\left[1+\left(\tfrac{99}{100}\right)^{k}+\ldots+\left(\tfrac{1}{100}\right)^{k}\right]=100^{k}\left(1+\varepsilon_k\right)\), where \(\varepsilon_k\to 0\).
Raising to the power \(\sin^2x=1/k\): \(\left[100^{k}(1+\varepsilon_k)\right]^{1/k}=100\cdot(1+\varepsilon_k)^{1/k}\).
Since \(0\le \varepsilon_k\to 0\), \((1+\varepsilon_k)^{1/k}\to 1\), so the required limit equals \(100\times 1=100\).
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