Why $\boldsymbol{\int_{-\infty}^0 \frac{dx}{3-4x}}$ does not exist?

Question

Can anyone explain why $\int_{-\infty}^0 \frac{dx}{3-4x}$ does not exist? I know this will get $$-0.25 \lim_{t\to -\infty}(\ln (3-0)-\ln(3-4t))$$ As I know $$\ln(3+((-4)(-\infty)))=\ln(3+\infty)=\infty$$

Answer 1

You have $${1\over 3 - 4x}\sim -{1\over4x}$$ as $x\to-\infty$. This last quantity does not integrate [finitely] at $-\infty$.

Answer 2

Make the substitution $x=-t$, and you arrive at harmonic series-like sum, so it diverges.