Well yes, I think what you meant is correct, but not what you wrote.
The first equality is only true if both limits exist, are finite and the denominator is non-zero, which is not true here. Then the second equality doesn't make sense, as $x$ is a bound variable and is not defined outside the integral, hence leading to the inconsistency that @QtizedQ pointed out. Finally, it should be clear that $\lim_{\Delta x\to 0}\int_0^{\Delta x}dx=0$ and so, in your last equality, you're dividing both the numerator and denominator by $0$.
However, what is true (assuming that $f$ and $g$ are integrable with $g$ non-zero) is that
$$\lim_{\Delta x\to0}\frac{\int_{x_1}^{x_1+\Delta x}f(x')dx'}{\int_{x_2}^{x_2+\Delta x}g(x')dx'}=\frac{f(x_1)}{g(x_2)}$$
You can show this pretty easily, using the Fundamental Theorem of Calculus. For that, let
$$F(x):=\int_0^xf(x)dx\quad\text{and}\quad G(x):=\int_0^xg(x)dx$$
Then
$$\lim_{\Delta x\to0}\frac{\int_{x_1}^{x_1+\Delta x}f(x')dx'}{\int_{x_2}^{x_2+\Delta x}g(x')dx'}=\lim_{\Delta x\to0}\frac{\frac{\int_{x_1}^{x_1+\Delta x}f(x')dx'}{\Delta x}}{\frac{\int_{x_2}^{x_2+\Delta x}g(x')dx'}{\Delta x}}=\lim_{\Delta x\to0}\frac{\frac{F(x_1+\Delta x)-F(x_1)}{\Delta x}}{\frac{G(x_2+\Delta x)-G(x_2)}{\Delta x}}=\frac{F'(x_1)}{G'(x_2)}=\frac{f(x_1)}{g(x_2)}$$
where the last equality comes from the Fundamental Theorem of Calculus.
As for the multivariable case, you can just apply the above result successively. If $\Delta V$ is a neighborhood of $(x,y,z)$, you can take the limit where its volume goes to $0$:
\begin{align}
\lim_{\text{vol}\left(\Delta V\right)\to0}\frac{\int_{\Delta V}f(x',y',z')dV}{\int_{\Delta V}g(x',y',z')dV}&=\lim_{\Delta x\to0, \Delta y\to0, \Delta z\to0}\frac{\int_x^{x+\Delta x}\int_y^{y+\Delta y}\int_z^{z+\Delta z}f(x',y',z')dx'dy'dz'}{\int_x^{x+\Delta x}\int_y^{y+\Delta y}\int_z^{z+\Delta z}g(x',y',z')dx'dy'dz'}\\
&=\lim_{\Delta x\to0, \Delta y\to0}\frac{\int_x^{x+\Delta x}\int_y^{y+\Delta y}f(x',y',z)dx'dy'}{\int_x^{x+\Delta x}\int_y^{y+\Delta y}g(x',y',z)dx'dy'}\\
&=\lim_{\Delta x\to0}\frac{\int_x^{x+\Delta x}f(x',y,z)dx'}{\int_x^{x+\Delta x}g(x',y,z)dx'}=\\
&=\frac{f(x,y,z)}{g(x,y,z)}
\end{align}
For a counterexample, take $f(x) = \cos x$, $a = 0$, $b = \infty$, and $\beta_n = n\pi$.
We have $\beta_n \to \infty$, and
$$\lim_{n \to \infty}\int_a^{\beta_n}f(x) \, dx = \lim_{n \to \infty}\int_0^{n\pi} \cos x\,dx = \lim_{n \to \infty}(\sin n\pi - \sin 0) = 0$$
However, the improper integral $\displaystyle\int_0^\infty \cos x \, dx$ does not converge, since $\underset{\beta \to \infty}\lim \sin \beta$ does not exist.
If $f \geqslant 0$, then $F(\beta)$ is nonnegative and nondecreasing. Suppose there exists a sequence $\beta_n$ where $\beta_n \to \infty$ and
$$\lim_{n \to \infty} F(\beta_n) = \lim_{n \to \infty}\int_a^{\beta_n} f(x) \, dx = I$$
For any $\epsilon > 0$, there exists $N$ such that $|F(\beta_n) - I| < \epsilon$ for all $n \geqslant N$. If $\beta > \beta_N$, then since $\beta_n \to \infty$, there exists $m > N$ such that $F(\beta_N) \leqslant F(\beta) \leqslant F(\beta_m),$ and
$$-\epsilon < F(\beta_N) - I < F(\beta)-I < F(\beta_m)- I < \epsilon$$
Thus,
$$\lim_{\beta \to \infty} F(\beta) = \lim_{\beta \to \infty}\int_a^{\beta} f(x) \, dx = I$$
Best Answer
Since $x^{-1/2}$ is undefined at $0$, the integral $\displaystyle\int_{-1}^0x^{-1/2}\,\mathrm dx$ doesn't exist in the sense of Riemann integration. And, by definition, we have$$\int_{-1}^0x^{-1/2}\,\mathrm dx=\lim_{t\to0^-}\int_{-1}^tx^{-1/2}\,\mathrm dx.$$That is, by definition $\displaystyle\int_{-1}^0x^{-1/2}\,\mathrm dx$ means $\displaystyle\lim_{t\to0^-}\int_{-1}^tx^{-1/2}\,\mathrm dx$.