will DTFT be equivalent of CTFT when the sample interval goes to zero

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Give an continuous natural signal x(t), then the CTFT of x(t) is $$X(\omega)=\int_{-\infty}^{+\infty} x(t) e^{-j \omega t} dt $$ The DTFT of x(t) is $$X(\Omega)= \sum_{n=-\infty}^{+\infty}x[n] e^{-j \Omega n}$$ As far as I know, $$Y(\omega)= X(\Omega) |_{\Omega=\omega T}$$ Where $$y(t)= \sum_{n=-\infty}^{+\infty} x(t)\delta(t-nT)$$ And T is the sample interval.

Will Y(ω) equal to X(ω) when T tends to zero? Intuitively, y(t) may equal to x(t) if T tends to zero, but how to understand it in a math way?

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It is easier to see if your signal is periodic with period $T$. Then your integrals should extend from $0$ to $T$. The sum for the DTFT should be multiplied by the time between samples. At that point it becomes just an approximation to the Riemann sum representing the integral. As the time between samples becomes smaller and smaller you approximate the integral better and better. That is the linkage you are looking for.