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?