Definition:
Suppose that $ f:[a,b] \rightarrow \mathbb{R} $ is continuous on $(a, b)$ and has singularities at $a$ and $b$.
If $\lim_{\epsilon \to 0+, \epsilon' \to 0+} \int_{a+\epsilon}^{b-\epsilon'} f = A$ for some real number $A$, then we define $\int_{a}^{b} f := A$.
What is the definition of the following formula?
$$\lim_{\epsilon \to 0+, \epsilon' \to 0+} \int_{a+\epsilon}^{b-\epsilon'} f = A.$$
Since $f$ has singularities at $a$ and $b$, the definite integral $\int_a^b f$ is undefined. But for sufficiently small $\epsilon$ and $\epsilon'$, $f$ is continuous on $[a+\epsilon,b-\epsilon']$. In those situations, $\int_{a+\epsilon}^{b-\epsilon'} f$ is defined. We can ask if there is a limit of that integral over the “truncated” closed interval as it grows closer to the full interval $[a,b]$. That is what is meant by the expression $\lim_{\epsilon\to 0,\epsilon'\to 0} \int_{a+\epsilon}^{b-\epsilon'} f$. If that limit exists (and is a real number $A$), we define the improper integral $\int_a^b f$ to be that limit.
To be more precise, $\int_a^b f = A$ if for every $\epsilon > 0$ there exist $\delta$ and $\delta'>0$ such that $$ \left|\int_{a+\eta}^{b - \eta'} f - A \right| < \epsilon $$ whenever $\eta$ and $\eta'$ satisfy $0 < \eta < \delta$ and $0 < \eta' < \delta'$.
You asked if this limit can be reduced to $\lim_{\epsilon\to 0} \int_{a-\epsilon}^{b+\epsilon}$. Consider the function $$ f(x) = \begin{cases} \frac{1}{x+1}+ \frac{1}{x-1} & x \neq \pm 1 \\ 0 & x = \pm1 \end{cases} $$ Then $f$ is defined on $[-1,1]$, continuous on $(-1,1)$ and has singularities at $1$ and $-1$. The indefinite integral is $$ \int\left(\frac{1}{x-1} + \frac{1}{x+1}\right)\,dx = \ln|x+1| + \ln|x-1| + C $$ So for any $\epsilon$ and $\epsilon'$ with $-1+\epsilon < 1 - \epsilon'$, we have \begin{align*} \int_{-1+\epsilon}^{1 - \epsilon'} f &= \ln(2-\epsilon') + \ln \epsilon' - \ln \epsilon - \ln (2-\epsilon) \end{align*} As $\epsilon$ and $\epsilon'$ tend to $0$, the first and last terms tend to $\ln 2$ and $-\ln2$, so they cancel. But what of $\ln \epsilon' - \ln \epsilon$? This is equal to $\ln\frac{\epsilon'}{\epsilon}$. If we restrict $\epsilon'$ to be equal to $\epsilon$, this ratio is $1$ and the definite integral is $0$. But if we restrict $\epsilon'$ to be a different multiple of $\epsilon$, say $\epsilon' = \alpha\epsilon$ for some $\alpha > 0$, then $\ln\frac{\epsilon'}{\epsilon} = \ln \alpha$. In summary, $$ \lim_{\epsilon\to 0}\int_{-1+\epsilon}^{1-\epsilon} f = 0 $$ but $$ \lim_{\epsilon\to 0,\epsilon'\to 0}\int_{-1+\epsilon}^{1-\epsilon'} f \text{ does not exist} $$