Why is the improper integral $\int^0_{-\infty}{1\over 3-4x}dx$ divergent?

Question

Take the following:

$$\int^0_{-\infty}{1\over 3-4x}dx$$

Substituting $t$ for $-\infty$, we can replace the above with

$$\left.\lim_{t\to-\infty}-{1\over4}\ln(3-4x)\right\rbrack^0_{t}\ ,$$

which evaluates to

$$\lim_{t\to-\infty}-{1\over4}(\ln3-\ln(3-4t))$$

$$ = -{1\over4}\ln3 + {1\over4}\ln(3-4t)$$

$$ = -{1\over4}\ln3 + \infty$$

According to my solutions manual, this integral is divergent. However, my understanding is that divergence with limits are only true when the entire limit evaluates to $\infty$. Given this understanding, or lack thereof, how exactly is it that this limit is divergent?

Answer 1

You've answered the question yourself. Just consider what the sum of $-\frac{1}{4}ln3 + {\infty}$ really is. You've clearly proven this is divergent.

Answer 2

If a function or integral increases without bound (that means $\to +\infty$) then adding or subtracting any constant will still result in that function increasing without bound. You might want to try to prove that to yourself using the definition of functions increasing without bound, in this case:

$\forall M > 0: \exists N < 0: t < N \implies -\frac 1 4 \ln 3 + \frac 1 4 \ln (3-4t) > M$

for $M$ sufficiently large.

Take $M' = M + \frac 1 4 \ln 3$, and so on.