Zero padding DFT intuition
Say a discrete signal $x(n)$ and its Z-transform $X(z)$, discrete time Fourier transform $X(e^{j\omega})$, and discrete Fourier transform $X(k)$. The relationships between DFT and ZT / DTFT are respectively:
$$
X(k) = X(z)|_{z=e^{j2\pi k/N}} \tag{1.1}
$$
$$
X(k) = X(e^{j\omega})|_{\omega=2\pi k/N} \tag{1.2}
$$
Since the original signal $x(n)$ can be perfectly reconstructed by IDFT, yes you can certainly obtain DTFT from DFT. However, you can't get DTFT by padding zeros as long as possible but by a technique called frequency domain interpolation. First is reconstruction of ZT from DFT:
$$
X(z) = \sum_{n=0}^{N-1}x(n)z^{-n}=\sum_{n=0}^{N-1}\Big[\frac{1}{N}\sum_{k=0}^{N-1}X(k)W_N^{-nk} \Big]z^{-n} \\ =  \frac{1}{N}\sum_{k=0}^{N-1}X(k)\Big[ \sum_{n=0}^{N-1}W_N^{-nk}z^{-n} \Big] \\ = \frac{1}{N}\sum_{k=0}^{N-1}X(k)\frac{1-W_N^{-Nk}z^{-N}}{1-W_N^{-k}z^{-1}} = \frac{1-z^{-N}}{N} \sum_{k=0}^{N-1}\frac{X(k)}{1-W_N^{-k}z^{-1}} \tag{2}
$$
where $W_N^{nk} = e^{j\frac{2\pi}{N}nk}$. Define the interpolation function
$$\varPhi_k(z) = \frac{1}{N}\frac{1-z^{-N}}{1-W_N^{-k}z^{-1}} = \frac{z^N-1}{Nz^{N-1}(z-W_N^{-k})} \tag{3}$$
and $X(z)$ can be written as
$$
X(z) = \sum_{k=0}^{N-1}X(k)\varPhi_k(z) \tag{4}
$$
Now let $z = e^{j\omega}$ we can derive that
$$
X(e^{j\omega}) = \sum_{k=0}^{N-1}X(k)\varPhi_k(e^{j\omega}) \tag{5}
$$
and the interpolation function
$$
\varPhi_k(e^{j\omega}) = \frac{1-e^{-j\omega N}}{N(1-W_N^{-k}e^{-j\omega})} = \frac{1-e^{-j\omega N}}{N(1-e^{-j(\omega-2\pi k/N)})}\\ = \frac{1}{N}\frac{\sin(N\omega/2)}{\sin\big[(\omega-2\pi k/N)/2\big]} e^{-j[\omega(N-1)/2 + k\pi/N]} \\
=\frac{1}{N}\frac{\sin(N\omega/2)}{\sin\big[(\omega-2\pi k/N)/2\big]} e^{jk\pi(N-1)/N} e^{-j(N-1)\omega/2} \tag{6}
$$
Let's write it in a more convinient form as
$$
\varPhi_k(e^{j\omega}) = \varPhi(\omega - 2\pi k/N) \tag{7}
$$
where $\varPhi(\omega)$ is the Fourier transform of a $N$-point rectangle window divided by $N$
$$
\varPhi(\omega) = \frac{1}{N} \frac{\sin(\omega N/2)}{\sin(\omega/2)} e^{-j(N-1)\omega/2} \tag{8}
$$
Finally, we get the equation of reconstructing DTFT by interpolating DFT
$$
X(e^{j\omega}) = \sum_{k=0}^{N-1}X(k)\varPhi(\omega-2\pi k/N) \tag{9}
$$