$$\displaystyle\underset{m\to +\infty }{\mathop{\lim }}\,\left[ \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\int_{0}^{x}{{{\left( \underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1-\frac{1}{n} \right)}^{n}} \right)}^{{{t}^{m}}}}t\text{d}t} \right]$$
2026-03-29 07:22:24.1774768944
A hard limit with integral sign
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$$\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1-\frac{1}{n} \right)}^{n}}=e^{-1}$$ and $$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\int_{0}^{x}{{e^{-t^m}}t\text{d}t}=0,$$ so the answer is $0$.