Bipolar junction transistors have no input resistance. 

An input resistance can be defined for _ports_ of n-port networks, such as the common-emitter configuration you've shown, with the two terminals connected to \$V_{in}\$ being one such port. If you connect an ideal voltage source \$V_{in}\$ to such a port, resulting in a current flow \$I_{in}\$, the input resistance is _defined_ to be 

$$ R_{in} := \frac{V_{in}}{I_{in}}.$$

Applied to your common-emitter circuit, we get \$ R_{in} = V_{in} / I_B \$. 

For DC inputs, \$R_{in}\$ varies strongly with \$V_{in}\$ (in this case) - the input resistance will be large for voltages below the BE-diode's forward voltage and very low for voltages above the forward voltage. You call this the _large signal_ input resistance.

When used as an amplifier, one would _bias_ the BJT suitably with a DC voltage to put it into the linear region and superimpose a small AC signal to be amplified. In this case, the _small signal_ input resistance \$r_{in}\$ is of interest. You can calculate \$r_{in}\$ from a small-signal model of a BJT and the various transistor parameters that can depend on the biasing conditions. Any textbook on BJTs or amplifier circuits should cover that.


Regarding the (common-emitter) current gain \$\beta\$: it is roughly constant only in the linear region of the BJT. (If it were not, there would be no linear region, so this is just a tautology.)