Solution to second order partial differential equation Suppose $Q$ is the solution to $Q_{tt}-Q_{xx} =0$. 
Let $\bar{Q}$ be the solution to $Q_{tt}-Q_{xx}+k Q_t=0$
Is there any known relation between these two solution, given that in the limit of $k$ going to zero, both solution should approach each other?
 A: In order to avoid confusion two different notations $P(x,t)$ and $Q(x,t)$ will be used :
$P$ is the solution to $\quad P_{tt}-P_{xx} =0$. 
$Q$ is the solution to $\quad Q_{tt}-Q_{xx}+k Q_t=0$
The general solution expressed on the form of integral is :
$$Q(x,t)=\int_{\forall \lambda} F(\lambda)e^{\lambda t +\frac12\left(k+\sqrt{k^2+4\lambda^2} \right)x }d\lambda +\int_{\forall \lambda} G(\lambda)e^{\lambda t +\frac12\left(k-\sqrt{k^2+4\lambda^2} \right)x }d\lambda $$
$F(\lambda)$ and $G(\lambda)$ are arbitrary functions to be determined according to some boundary conditions (not mentioned in the wording of the question).
In case of $k=0$ the general solution is the preceding equation which is reduced to :
$$P(x,t)=\int_{\forall \lambda} f(\lambda)e^{\lambda (t+x) }d\lambda + \int_{\forall \lambda} g(\lambda)e^{\lambda (t-x) }d\lambda$$
$f(\lambda)$ and $g(\lambda)$ are arbitrary functions to be determined according to some boundary conditions. There is no reason for $f=F$ and $g=G$.
Note that one can write equivalently
$$P(x,t)=\Phi(t+x)+\Gamma(t-x)$$
where $\Phi$ and $\Gamma$ are arbitrary functions.
The question raised by the OP is if whether or not a general relationship exist between $P(x,t)$ and $Q(x,t)$.
Since both $P(x,t)$ and $Q(x,t)$ depends from boundary conditions it is doubtful that such a relationship exists because they are an infinity of possible different boundary conditions.
Supposing that the question be less general, but raised for only a particular boundary condition common to the two PDEs, it is difficult to give a definitive answer without the particular boundary condition be specified. 
APPROXIMATE SOLUTION for $k$ close to $0.$ :
$\lambda t +\frac12\left(k\pm\sqrt{k^2+4\lambda^2} \right)x \simeq\lambda t +\frac12\left(k\pm 2\lambda \right)x $
$$Q(x,t)\simeq\int_{\forall \lambda} F(\lambda)e^{\lambda t +\frac12\left(k+2\lambda \right)x }d\lambda +\int_{\forall \lambda} G(\lambda)e^{\lambda t +\frac12\left(k-2\lambda \right)x }d\lambda $$
$$Q(x,t)\simeq\int_{\forall \lambda} F(\lambda)e^{\lambda (t+x) +\frac{k}{2} x }d\lambda +\int_{\forall \lambda} G(\lambda)e^{\lambda (t-x) +\frac{k}{2}x }d\lambda $$
$$Q(x,t)\simeq e^{\frac{k}{2} x}\int_{\forall \lambda} F(\lambda)e^{\lambda (t+x) }d\lambda +e^{\frac{k}{2} x}\int_{\forall \lambda} G(\lambda)e^{\lambda (t-x)}d\lambda $$
$$Q(x,t)\simeq e^{\frac{k}{2} x}\big( \Phi(t+x)+\Gamma(t-x)\big)+\epsilon(x,t)\qquad k\:\:\text{small}$$
$$Q(x,t)\simeq e^{\frac{k}{2} x}P(x,t)+\epsilon(x,t)$$
The magnitude of the deviation $\epsilon(x,t)$ is proportional to $k^2$.
