Limit of the $\sin^2$ function in the derivation of Fermi's golden rule In the derivation of Fermi's golden rule one typically arrives at an expression of the form
$$
\frac{\sin^2(\omega t)}{\omega^2}
$$
which is then converted to 
$$
\pi t\delta(\omega).
$$
I cannot follow this step. I know the following identity 
$$
\delta (\omega) = \lim_{t\rightarrow \infty}\frac{\sin^2(\omega t)}{\pi |t|\omega^2}
$$
from which i would assume that one extends the first expression by $\frac{t}{t}$ and then does the limit. But how can you pull the $t$ out of the limit ? Is this rigorous or is this an approximation ?
I think it should be like this
$$
\lim_{t\rightarrow\infty} \frac{t}{t}\frac{\sin^2(\omega t)}{\omega^2} \not = \pi t\delta(\omega)
$$
and the equation should be 
$$
\lim_{t\rightarrow\infty} \frac{t}{t}\frac{\sin^2(\omega t)}{\omega^2} = \pi \delta(\omega) \lim_{t\rightarrow \infty}t.
$$
Am i wrong in the above equations ? Otherwise i don't see how Fermi's goldene rule could ever work since we assume at one time that $t$ is so large that we can approximate a function in the limit that $t$ goes to infinity while on the other hand $t$ has to be small such that pertubation theory of first order is accurate. These conditions seem to contradict each other but in every book i find this step. I haven't found any satisfactory answer so far regarding this step. I know the general conditins for pertubation theory but i find the form with the dirac delta function nonsensical. I assume that i go wrong at some point since no one ever brings this point up, please point out my error if i did something wrong.
 A: *

*In the comments it became clear that OP is trying to understand the formula
$$ \lim_{t\to\infty} F\left(t,\frac{E-E_i}{\hbar}\right)~=~\pi t~\delta\left(\frac{E-E_i}{2\hbar}\right)~=~2\pi\hbar t~\delta(E-E_i) , \tag{C-32} $$ 
taken from Ref. 1, where 
$$ F(t,\omega)~=~\left[\frac{\sin(\omega t/2)}{\omega/2}\right]^2. \tag{C-7} $$ 

*It is clear that eq. (C-32) does not make sense as an ordinary limit, since the rhs. should then be independent of $t$. (It makes sense as an asymptotic series, though.)

*However, if the goal is just to derive Fermi's golden rule, there is a short-cut: It turns out that the $F$-function (C-7) is proportional to the probability $P(t)$, which grows with time $t$. The trick is to instead consider the quotient $\frac{P(t)}{t}$. Then it becomes mathematically well-defined to consider the limit $\lim_{t\to\infty}\frac{P(t)}{t}$, i.e. 
$$ \lim_{t\to\infty} \frac{F\left(t,\frac{E-E_i}{\hbar}\right)}{t}~=~\pi ~\delta\left(\frac{E-E_i}{2\hbar}\right)~=~2\pi\hbar ~\delta(E-E_i) , \tag{C-32'} $$ 
cf. this Phys.SE post. This in turn implies that for large enough times $t$ the probability $P(t)$ grows proportionally with $t$, and that the transition rate 
$$ \frac{dP(t)}{dt}~=~\frac{P(t)}{t} \tag{*}$$
is given by the proportionality factor.

*For more details of the proof of Fermi's golden rule, see my Phys.SE answer here.
References: 


*

*C. Cohen-Tannoudji, B. Diu & F. Laloe, QM, Vol. 2, 1978; Chapter XIII, p. 1283-1302.

