Proving that $\lim\limits_{x \to 0^+}\log x\to-\infty$ and $\lim\limits_{x \to +\infty} \log x  \to +\infty $ using $\log x = \int_1^x \frac {dt} t$ Define $L:(0,\infty)\to\mathbb{R}$ by
$$L(x)=\int_{1}^{x}\frac{dt}{t}.$$
Show that $\lim_{x\to0^{+}}L(x)=-\infty$ and $\lim_{x\to\infty}L(x)=\infty$.
This is what I have done: We have that
$$\lim_{x\to0^{+}}L'(x)=\lim_{x\to0^{+}}\frac{1}{x}=\infty.$$
So, it must be the case that $\lim_{x\to0^{+}}L(x)=\pm\infty$. Moreover, we have that
$$L(x)=\int_{1}^{x}\frac{dt}{t}=-\int_{x}^{1}\frac{dt}{t}.$$
Hence, this implies that $\lim_{x\to0^{+}}L(x)=-\infty$.
For the other case, we have that
$$\lim_{x\to\infty}\int_{1}^{x}\frac{dt}{t}>1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$
But since this sum does not converge, it must be the case that $\lim_{x\to\infty}L(x)=\infty$.
What do you guys think?
 A: The derivative of $f$ having an infinite limit does not imply that $f$ has an infinite limit. Consider $\lim\limits_{x\rightarrow0^+}\sqrt x$.
But, here's a hint: 
You can show that  $\int_\delta^{2\delta} {1\over t}\,dt\ge\delta\cdot{1\over2\delta}= {1/2}$ for any positive $\delta$. This would imply that $\int_0^1{1\over t}\,dt$ diverges to $\infty$ (consider the integral over the intervals intervals $[1/4,1/2]$, $[1/8,1/4]$, $[1/16,1/8]$, $\ldots\,$). From this, it follows that $\lim\limits_{x\rightarrow0^+}L(x)=-\infty$.
You could also show, by considering the integral over the intervals $[1 ,2 ]$, $[2,4]$, $[4,8]$, $\ldots\,$
that $L(2^n)\ge n\cdot {1\over2}$. This would imply that $\lim\limits_{x\rightarrow\infty}L(x)=\infty$.
A: You can show that $\lim_{x\to\infty}L(x)=\infty$ first, then deduce that $\lim_{x\to0^{+}}L(x)=-\infty$.  But, as joriki mentioned, your estimate needs to be made more precise.  As a hint:
$$
\int_{1}^{x} \frac{dt}{t} = \int_{\lfloor x \rfloor}^{x} \frac{dt}{t} + \sum_{k=1}^{\lfloor x \rfloor - 1} \int_{k}^{k+1} \frac{dt}{t}.
$$
Then, once you've shown that $\lim_{x\to\infty}L(x)=\infty$, consider the substitution $t = 1/u$, giving
$$
\int_{1}^{x} \frac{dt}{t} = - \int_{1}^{\frac{1}{x}} \frac{du}{u}.
$$
You only need to do the work once due to the symmetry of the function $1/t$.  Integrating from $1$ to $\infty$ is basically the same as integrating from $0$ to $1$.

A: On your idea of proving that the sum $S(M)$ 
$$S(M)=\sum_{n=1}^M \int_n^{n+1}\frac {dt} t$$ you can argue as follows (make a graph, or use that $f$ is decreasing):
$$\frac{1}{n+1}<\int_n^{n+1}\frac {dt} t<\frac{1}{n}$$
$$\sum_{n=1}^M \frac{1}{n+1}<\sum_{n=1}^M \int_n^{n+1}\frac {dt} t<\sum_{n=1}^M \frac{1}{n}$$
$$\sum_{n=1}^M \frac{1}{n+1}<\int_1^{M+1}\frac {dt} t<\sum_{n=1}^M \frac{1}{n}$$
As you can see from other answers, the result for $\displaystyle \int_0^1 \frac{dt}t$ follows as a corollary.
