# Tensionless string in Nambu-Goto action

I am studying string theory from the book "String theory and M-theory", written by Becker, Becker and Schwartz. My question is:

We are taught that one of the advantages of introducing a field $$e(\tau)$$ into the action of the point particle, namely $$S_0=-m\int d\tau\sqrt{-\frac{dX^{\mu}}{d\tau}\frac{dX_{\mu}}{d\tau}} \rightarrow \tilde{S}_0=-m\int d\tau e(\tau)\Big[e^{-2}(\tau)\dot{X}^2-m^2\Big]$$ is to also include in the analysis the case in which the particle is massless. Furthermore, this procedure is repeated in the case of the string, which is a one dimensional object. We introduce the auxilliary metric $$h_{\alpha\beta}$$, such that we get rid of the square root in the Nambu-Goto action. If I were to extend this correspondence, I should claim that one of the advantages of introducing the auxilliary field is to cover the case of the tensionless string as well as the cases in which the string has non-zero tension. Is this true? If so, what does it mean for a string to be tensionless? Does that even make sense (to talk about a tensionless string)?

Consider the Nambu-Goto action for the relativistic string $$S_{NG} = -T\int d\tau d\sigma_{proper}$$ where the integral represent the proper area of the world sheet in spacetime.The coefficient $$T$$ is the string tension.
The observations is that the $$T \rightarrow0$$ limit makes $$S_{NG}$$ meaningless. Even more clear, notice that $$S$$-matrix elements (the string theory observables) are computed in perturbative string theory as a power series in $$\alpha^{\prime}=\frac{1}{2\pi T}$$, any sort of $$T \rightarrow0$$ limit is over perturbative control.
However, there are actually tensionless strings (known as self-dual strings) in string theory. The cavet is that their theory is currently unknown. The basic configuration arises as an $$M2$$-brane suspended over two parallel $$M5$$-branes.