Proving $\lim_{x \to 0} \, \int_1^x \frac{dt}{t} = -\infty$ I am looking for hints on how to prove $$
\lim_{x \to 0} \, f(x) = \lim_{x \to 0} \, \int_1^x \frac{dt}{t} = -\infty.
$$
I know that $f(x) = \log(x)$ and that $\lim_{x \to 0} \, \log(x) = -\infty$. However, I am interested in proving the statement by only appealing to properties of the integral: $\int_1^x \frac{dt}{t}$.
The fact that $1/x \to \infty$ as $x \to 0$ is not sufficient to prove the statement since the area under the curve could still be finite even if the integrand diverges.
Many thanks for any suggestions here.
 A: Make the substitution $s = \frac{1}{t}$. Making the substitution yields
$$\int_1^\infty \frac{1}{s} \, \mathrm{d}s.$$
To show that this integral doesn't exist, use the integral test for series. In particular, the function $\frac{1}{x}$ is dominates the function that takes the value $\frac{1}{n+1}$, for $n < x \le n + 1$. The integral of this function is divergent, as the area under each "pillar" is a term in the harmonic series, which is known to be divergent. Hence, the integral diverges.
A: By definition, for $0<x<1,$
$$\int_1^x \frac{dt}{t}=-\int_x^1 \frac{dt}{t}.$$
You want to show that the limit of this as $x\to 0^+$ equals $-\infty.$ In other words, you want to show
$$\tag 1  \int_0^1 \frac{dt}{t} = \infty.$$
Let $t=2s.$ Then $(1)$ equals
$$\tag 2 \int_0^{1/2} \frac{ds}{s}.$$
This is impossible unless both integrals equal $\infty.$ 
A: You should start off evaluating the integral. This would lead you to get $ln(x) - ln(1)$ and since $ln(1) = 0$, you're left with the limit of ln(x) as x approaches 0.
If you graph ln(x), you'll see that as x approaches 0, y approaches negative infinity.
This should prove your limit.
A: For $n$ a positive integer you have
$$\int_{2^{-n}}^{2^{-n-1}} \frac{1}{t} \; dt < \int_{2^{-n}}^{2^{-n-1}} \frac{1}{2^{-n}} \; dt = \left(\frac{1}{2^{n+1}} - \frac{1}{2^n}\right) 2^n = -\frac{1}{2}.$$
Then
$$\lim_{x\to 0} f(x) = \lim_{n\to \infty} \int_1^{2^{-n-1}} \frac{1}{t} \; dt = 
\lim_{n \to \infty} \sum_{k=0}^{n} \int_{2^{-k}}^{2^{-k-1}} \frac{1}{t} \; dt < \frac{-n}{2} = -\infty.$$ 
A: First, since $\frac{1}{x} > 0$ for $x > 0$, $f$ is strictly increasing; so either $\lim_{x\to 0^+} f(x) = -\infty$ or $\lim_{x\to 0^+} f(x) = L$ for some $L \in \mathbb{R}$.  We will prove that the last case leads to a contradiction.
So, suppose $\lim_{x \to 0^+} f(x) = L$, and choose $\epsilon > 0$ such that $f(\epsilon) < L + f(2)$.  Then:
$$ f(\epsilon/2) = \int_1^{\epsilon/2} \frac{dt}{t} = \int_{2}^{\epsilon} \frac{du/2}{u/2} = \int_{2}^{\epsilon} \frac{dt}{t} = f(\epsilon) - f(2) < L,$$
giving the desired contradiction (combined with the fact that $f$ is strictly increasing, so $f(x) > L$ for every $x > 0$).
(Now that I look again, I see that zhw. gave essentially the same solution; it's just that this one presents the argument a bit more rigorously.)
A: Thanks for the feedback all. I am going contribute my own answer as well which was inspired by @B. Goddard's answer.
We know that $$
\lim_{x\to^+ 0} \int_1^x \frac{dt}{t} = -\lim_{x\to^+ 0} \int_x^1 \frac{dt}{t}.
$$
Moreover, the integral $\lim_{x\to^+ 0} \int_x^1 \frac{dt}{t}$ is positively valued and lower bounded by the sums $$
S_N = \sum_{n=1}^N \frac{1}{N} \cdot \frac{1}{n/N} = \sum_{n=1}^N \frac{1}{n}
$$
where $1/N$ is the partition width of the Riemann sum and $1/(n/N)$ is the height of the $n$'th partition.
Since the Riemann integral must be the sup over all sums $S_N$ (assuming it exists), it must be at least as big as $$
\lim_{N\to\infty} \sum_{n=1}^N \frac{1}{n}
$$
which we know diverges.
I guess not so interesting after all...
