Apparend frequency when the source travels with the speed of sound Let's assume an object travels with the speed of sound towards a stationary observer in a straight line. Let's say the frequency emitted by the jet is $f,$ speed of sound is $u. $ and the speed of jet also is $u $. Then the apparent frequency the observer will receive (say $f_1$) is;
\begin{eqnarray}
f_1 = [u/(u-u)] \times f
\end{eqnarray}
Will $f_1$ tend to $\infty$?
And what if the jet travels with supersonic speed (say Mach 3) — will $f_1$ be negative?
 A: 
Will $f_{1}$ tend to $\infty$?

Kind of, but not in the way I think you are implying.  The acoustic wave generated by the moving source can steepen into a discontinuity known as a shock wave.  In Fourier space, a time discontinuity turns into an infinite number of frequencies.
If something moved at the local speed of sound and emitted a monotonic tone, an observer at rest may not hear it until the source passes.  The problem with the simplification of the Doppler effect discussed here is that the direction of the emission matters and it's only supposed to be applied for speeds much smaller than the speed of sound, $C_{s}$.  The full approximation is given by:
$$
f_{r} = \left( \frac{ C_{s} \pm v_{r} }{ C_{s} \pm v_{s} } \right) f_{s} \tag{0}
$$
where subscript $r$($s$) refers to the receiver(source) and the $v_{j}$ is the speed of the jth constituent with respect to an at rest medium.  In the numerator, the plus(minus) refers to the receiver moving toward(away from) the source.  In the denominator, the plus(minus) refers to the source moving away from(toward) the receiver.
So for a source moving with $v_{s} = C_{s}$ toward a receiver and the receiver is at rest, $v_{r} = 0$, Equation 0 reduces to:
$$
f_{r} = \left( \frac{ C_{s} }{ C_{s} - v_{s} } \right) f_{s} \tag{1}
$$
When the source is still moving toward the receiver, i.e., it has not yet past, the receiver will hear nothing as the sound waves have not yet been received.  Once the source has passed by the receiver, the receiver will hear sounds at half the frequency of the source because the minus in the denominator of Equation 1 turns into a plus after the source passes.

And what if the jet travels with supersonic speed (say Mach 3) — will $f_{1}$ be negative?

Again, kind of but not in the way I think you are implying.  It's not so much that there are negative frequencies that humans cannot hear, but rather that "negative" just refers to the mirrored image of the "positive" frequencies here.  If the emitted waves were circularly polarized, right-handed electromagnetic modes, they could be detected by a moving observer as left-handed modes.  This is one effect of a the "negative" to which I refer.
As an example, suppose the source moves at $v_{s} = 3 \ C_{s}$, then the receiver will hear nothing until the source has passed and what it does measure will be four times lower in frequency than the source.  Let's reverse this a little to make it easier to understand.
Suppose the source is stationary, $v_{s} = 0$, but the receiver moves toward it with $v_{r} = 3 \ C_{s}$.  The source is isotropically emitting sound waves so while the receiver moves toward it, the receiver will measure $f_{r}$ ~ 4$f_{s}$.  When the receiver passes the source and starts to receed, the measured frequency will be $f_{r}$ ~ -2$f_{s}$.  This does not mean the receiver measures nothing or anything special, it's just a "reverse ordered" version of the emission at twice its original frequency.  The receiver is overtaking the sound waves, so the phase fronts are received in reverse order to what they were emitted.  If the signal is something simple like a monotone, then a simple frequency shift is measured.  If the signal were a speech, the quick garbled version would be received.
