Continuity of a Logarithmic Integral Define $L:(0,\infty)\to\mathbb{R}$ by
$$L(x)=\int_{1}^{x}\frac{dt}{t}.$$
How can I show that $L$ is continuous on $(0,\infty)$?
I know that the definition of continuity states the following:

A function $f:A\to\mathbb{R}$ is continuous at a point $c\in A$ if, for all $\epsilon>0$, there exists a $\delta>0$ such that whenever $|x-c|<\delta$ (and $x\in A$), it follows that $|f(x)-f(c)|<\epsilon$.
If $f$ is continuous at every point in the domain $A$, then we say that $f$ is continuous on $A$.

I have tried applying this definition by letting $x,c\in(0,\infty)$. Then
$$\left|L(x)-L(c)\right|=\left|\int_{1}^{x}\frac{dt}{t}-\int_{1}^{c}\frac{dt}{t}\right|=\left|\int_{1}^{x/c}\frac{dt}{t}\right|=\left|L(x/c)\right|<\epsilon,$$
but no matter how hard I look, I cannot seem to find the connection between that and $|x-c|<\delta$, because these are integrals, and I cannot perform algebraic manipulations to reach the desired inequality.
Do you guys have any ideas?
 A: The proofs one can use depend on what has been proved so far. There are two alternatives. One could question you about whether various theorems have been proved, in your course, or one could guess. I will guess that only very basic properties of the definite integral have been proved rigorously. 
Let $c\in (0,\infty)$. We show that $L$ is continuous at $c$. That means that for any $\epsilon \gt 0$, we must produce (or at least show the existence of) a $\delta \gt 0$ such that if $|x-c| \lt \delta$, then $|L(x)-L(c)| \lt \epsilon$. Note that 
$$|L(x)-L(c)|=\left|\int_1^x\frac{dt}{t} -\int_1^c\frac{dt}{t}\right|=\left|\int_c^x\frac{dt}{t}\right|.$$
We estimate the last integral. If $\frac{1}{t} \lt M$ in the interval from $c$ to $x$ (but $x$ could be to the left of $c$), we have
$$\left|\int_c^x\frac{dt}{t}\right| \le |x-c|M.$$
We first of all make $\delta<c/2$. That makes sure that $t$ does not dip below $c-c/2=c/2$, so in the interval of integration, $\frac{1}{t}$ is bounded above by $M=\frac{2}{c}$. 
If we make that restriction on $\delta$, then if $|x-c| \le \delta$, we have
$$\left|\int_c^x\frac{dt}{t}\right| \le |x-c|M \le \delta\frac{2}{c}\tag{$\ast$}.$$
Make $\frac{2\delta}{c} \lt \epsilon$, that will do it! So let's make $\delta \lt c\epsilon/2$.  For definiteness, make $\delta=c\epsilon/5$, why not? That's not quite right, remember that we wanted $\delta<c/2$. So make $\delta$ equal to the minimum of $c/2$ and $c\epsilon/5$. This ties down everything. Part of it was overkill, as protection against the possibility that as a joke we are given $\epsilon=1000$. If we look only at small $\epsilon$, say less than $1$, then $\delta=c\epsilon/5$ is plenty good enough, indeed $\delta=c\epsilon/2$ is good enough.  
The paragraph above is of very slight importance. In essence, once we had the inequality $(\ast)$, it was clear that by appropriate choice of $\delta$ we could make $|L(x)-L(c)|$ as small as we wished.
Remark: The approach you tried also works, and the details are not all that different. It requires first proving the fact that the difference is $\int_1^{x/c}\frac{dt}{t}$, but I imagine that was done in your course. (That special fact would not be available if we had say $1/\sqrt{t}$ instead of $1/t$.) Then we can find an upper bound for the absolute value, of the shape $|x/c-1|M$, where $M$ is as in the solution above. Finally, we need to estimate $|x/c-1$ in terms of $|x-c|$, that part is easy. 
A: $f$ be integrable on the $[a,b]$, then $F=\int_{a}^{x}f(t)dt$ is continuous apply this result to prove L is continuous  
A: To show continuity on $(0,\infty)$, it is sufficient to show that $L$ is continuous on the interval $(a,\infty)$ for any $a>0$. The advantage of doing this is that you can easily bound $t \mapsto \frac{1}{t}$ on the interval $(a, \infty)$ by $\frac{1}{a}$.
The we have, for $x,y \in (a, \infty)$,
$$| L(x)-L(y) | = |\int_y^x \frac{1}{t} dt | \leq \frac{1}{a} | x-y|.$$
Choosing $\delta < a \epsilon$, then you have the required result, ie, if $|x-y| < \delta$, then $| L(x)-L(y) | \leq \frac{1}{a} 
\delta < \epsilon$.
Since $a>0$ was arbitrary, you can conclude that $L$ is continuous on $(0,\infty)$.
A: Take a close interval containing c, there the function f(t)=1/t is bounded.
 Then use  monotony of integral between c to x.
(For x sufficient close to c you are working in the close interval you took).
