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Specifically, I ran into trouble when trying to solve $$\lim\limits_{n \to \infty} \int_1^n x^{-n} \, dx.$$
The integral is very easy to solve but what next? The limit $\displaystyle\lim\limits_{n\to \infty} \frac{n^{-n+1}-1}{-n+1}$ looks like $\displaystyle\lim\limits \frac{\infty^{-\infty}}{-\infty}$ which I'm not sure how to solve. Can I apply L'Hopital Rule here?
Hint: Let $t=\dfrac1x$ , and use the fact that $\displaystyle\lim_{a\to\infty}x^a=0$ when $x\in(0,1)$.
As you wrote, you get $$ \lim_{n\to \infty} \frac{n^{-n+1}-1}{-n+1}. $$ Note that there is a minus sign in the exponent, so this should not look like $\frac{\infty^\infty}{-\infty}$ but something simpler: just find the limits of the numerator and denominator separately, and you should see the result.
Exponentiation (of extended real numbers) is defined and continuous there, so you can just plug things in, using $(+\infty)^{-\infty} = 0$.
Similarly, $(+\infty)^{+\infty} = +\infty$
inf^(-inf)/-inf = 1/inf*(-inf) = -1 / inf^2 --> 0
$$\frac{\infty^{-\infty}}{-\infty} = \frac1{\infty\cdot(-\infty)} = -\frac1{\infty^2} \to 0$$
