How can I find the Fourier transform of constant value like $1$. The textbook told me that $\mathbb F[1] = \delta(f)$ and $\mathbb F[\delta(t)]=1$.

It is easy to prove that $\mathbb F[\delta(t)] = 1$.
$$
\mathbb F[\delta(t)] = \int_{-\infty}^\infty \delta(t)e^{-j2\pi ft}dt = e^{-j2\pi ft}|_{t=0}=1
$$

However, I failed to prove that $\mathbb F[1] = \delta(f)$.
I don't want to prove this using inverse Fourier transform like the following.
$$
\mathbb F^{-1}[\delta(f)] = \int_{-\infty}^\infty \delta(t)e^{j2\pi ft}dt = e^{j2\pi ft}|_{t=0}=1\\
\therefore \mathbb F[1] = \delta(f)
$$

My trial:
\begin{align}
\mathbb F[1] &= \lim_{k\to\infty}\int_{-k}^k 1\cdot e^{-j2\pi ft}dt\\\\
&=\lim_{k\to\infty}\left[\frac{e^{-j2\pi ft}}{-j2\pi f}\right]^k_{-k}\\\\
&=\lim_{k\to\infty} \frac {-j2\sin(2\pi fk)}{-j2\pi f}\\\\
&=\lim_{k\to\infty} 2k\cdot\frac{\sin(2\pi fk)}{2\pi fk}\\\\
&=\lim_{k\to\infty} 2k\cdot\mbox{sinc}(2fk)\\\\
&=\left\{
\begin{array}{ccc}
\displaystyle\lim_{k\to\infty}{2k}=\infty &&f=0\\
\displaystyle\lim_{k\to\infty}{2k}\cdot\mbox{sinc}(2fk)=? &&\mbox{otherwise}
\end{array}
\right.
\end{align}
Please someone tell me some hints or let me know where I have to head for. Thank you.
 A: Let me try to give an intuitive understanding of the limit you need for proving the identity in your question. Assuming you have a signal processing background (judging from other questions of yours), you probably know (or can easily look up) that the impulse response of an ideal low pass filter with cut-off (angular) frequency $\Omega$ is given by
$$h(t)=\frac{\sin(\Omega t)}{\pi t}\tag{1}$$
Now imagine what happens when the cut-off frequency $\Omega$ is increased: the impulse response gets more and more narrow, and in the limit you don't have a low pass filter anymore but no filtering at all (because $\Omega\rightarrow\infty$). This means
$$\lim_{\Omega\rightarrow\infty}\frac{\sin(\Omega t)}{\pi t}=\delta(t)\tag{2}$$
because convolution with $\delta(t)$ is the identity operation (no filtering).
In the problem of your question you have the limit
$$\lim_{k\rightarrow\infty}\frac{2\sin(\omega k)}{\omega}\tag{3}$$
with $\omega=2\pi f$. Comparing this limit with the limit in $(2)$ with the correspondences $\omega\leftrightarrow t$ and $k\leftrightarrow\Omega$, we see that the limit in $(3)$ must equal $2\pi\delta(\omega)=\delta(f)$, (with $\omega=2\pi f$).
