$\infty -\infty = \ ?$

The answer is in the topic of the question. lol

$\lim_{t \to \infty} \left ( \int_{\frac{1}{t}}^1{\frac{1}{x}\,dx} - \int_1^t{\frac{1}{x}\,dx} \right )$

$= \displaystyle \lim_{t \to \infty} \left(\ln(x)\bigg|_{\frac{1}{t}}^1 - \ln(x)\bigg|_1^t \right) \\ = \displaystyle \lim_{t \to \infty} \left(\ln(1)-\ln\left(\frac{1}{t}\right) - \ln(t) + \ln(1) \right) \\ = \displaystyle \lim_{t \to \infty} \left (\ln(t) - \ln(t)\right) \\ \implies \displaystyle \lim_{t \to \infty} (0) = \boxed{0}$