finding limit cycles and analysis the behavior of a nonlinear system Consider the following nonlinear system
$\dot{x}=1-e^{y}$ $\quad$
$\dot{y}=1-e^{y-x}$
This system is asymptotically stable; however, when we simulate the trajectories of this system, there exist no trajectories when $y$ goes to infinity. I am going to investigate the behavior of this system for when $x\rightarrow -\infty$ and $y\rightarrow \infty$.
In addition, another question is that in your view this system has limit cycle
 A: Though I find this question to be slightly vague, I think a useful picture of the dynamics of this system may be had via the application of the usual, standard methods and techniques.  We are presented with
$\dot x = X(x, y) = 1 - e^y \tag 1$
and
$\dot y = Y(x, y) = 1 - e^{y - x}; \tag 2$
we first find the critical points or zeroes of the vector field $(X, Y)$:
$1 - e^y = X = 0 \Longleftrightarrow e^y = 1 \Longleftrightarrow y = 0; \tag 3$
then (2) yields
$1 - e^{-x} = Y = 0 \Longleftrightarrow e^{-x} = 1 \Longleftrightarrow x = 0; \tag 4$
we thus see that the only critical point of the system is
$(x, y) = (0, 0); \tag 5$
the Jacobean of the system at the general point $(x, y)$ is
$J(x, y) = \begin{bmatrix} X_x & X_y \\ Y_x & Y_y \end{bmatrix}, \tag 6$
where
$X_x = \dfrac{\partial X}{\partial x} \tag 7$
and so forth;  we have
$J(x, y) = \begin{bmatrix} 0 & -e^y \\ e^{y - x} & -e^{y - x} \end{bmatrix}, \tag 8$
at the specific point $(0, 0)$ this becomes
$J(0, 0) = \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix}, \tag 9$
we compute the eigenvalues of $J(0, 0)$ by finding its characteristic polynomial
$\det(J(0, 0) - \lambda I) = \det \left (\begin{bmatrix} -\lambda & -1 \\ 1 & -1-\lambda \end{bmatrix} \right )$
$= \lambda(\lambda + 1) + 1 = \lambda^2 + \lambda + 1; \tag{10}$
the roots of
$\lambda^2 + \lambda + 1 = 0 \tag{11}$
are given by the quadratic formula:
$\lambda_\pm = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2}$
$= \dfrac{-1 \pm \sqrt{-3}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt 3}{2} i; \tag{12}$
since these eigenvalues have negative real part, we may invoke the  stable manifold theorem to infer $(0, 0)$ is a stable spiral equilibrium of (1), (2); that is, any trajectory of (1)-(2) initialized sufficiently near $(0, 0)$ will approach the origin taking a spiral path;  indeed, the linearized system is consistent with this, since
$(-y, x) \cdot J(0, 0) (x, y) = (-y, x) \cdot (-y, x -y)$
$= y^2 + x^2 -xy  > 0 \; \text{if} \; (x, y) \ne (0, 0), \tag{13}$
so we see that the solutions spiral in a counter-clockwise direction, the direction in which $(-y, x)$ points; likewise, evaluating
$(x, y) \cdot J(0, 0) (x, y) = (x, y) \cdot (-y, x -y)$
$= -xy + xy -y^2 = -y^2 < 0 \; \text{if} \; y \ne 0, \tag{14}$
we confirm that the solution moves strictly inward except possibly when crossing the $x$-axis, again for $(x, y)$ in a sufficiently small neighborhood of the origin.
The above argument yields a local description of the qualitative dynamics of (1)-(2) near $(0, 0)$, which indeed exhibit asymptotic stability there; we next turn to 
the question of limit cycles and periodic orbits in general.  In fact, this system has no periodic trajectories, as a consequence of the Bendixson-Dulac theorem, which asserts there can be no such orbit in any region in which 
$\nabla \cdot (X, Y) = X_x + Y_y \tag{15}$
is of fixed sign, except perhaps on a set of measure zero.  (The actual theorem is somewhat more general, but this will suffice for the present purposes.)  In light of (1), (2) we have
$X_x + Y_y = -e^{y - x} < 0 \tag{16}$
everywhere; thus by the cited theorem there are no periodic trajectories, hence in particular no limit cycles. 
In the above discussion we have exploited the standard, basic methods of the theory of non-linear ODEs to demonstrate certain fundamental facts concerning the solution space of the system (1)-(2); however, our story is far from complete.  We would like to know, for example, how the system behaves far from the origin, or if the origin is a global attractor; but these questions seem to require a much deeper and more lengthy analysis, and so will be not be addressed here.
