Determine the marginal distributions of $(T_1, T_2)$ 
A random variable $(T_1, T_2)$ is called two-dimensional exponentially
  distributed with parameters $(\lambda_1, \lambda_2, \lambda_3)$, if
  the distribution function has the form
$$F(t_1,t_2) = \left\{\begin{matrix}
1-e^{-(\lambda_1+\lambda_3)t_1}-e^{-(\lambda_2+\lambda_3)t_2}+e^{-\lambda_1t_1-\lambda_2t_2-\lambda_3 \text{max}(t_1,t_2)}, \text{ if } t_1>0, t_2>0\\  0
 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{
else } \end{matrix}\right.$$
Determine the marginal distributions of $(T_1,T_2)$

I never do such task that's why I find this example to see how it work correct so I understand better and do it correct in test hopefully.
When I understand correct, I need to calculate limit to $\infty$ of the distribution function for $t_1$ and seperately for $t_2$ then I have the marginal distributions of $(T_1,T_2)$.
$$\lim_{t_1 \rightarrow \infty}f(t_1,t_2) = 1-e^{-\infty}-e^{-(\lambda_2+\lambda_3)t_2}+e^{-\infty} = 1+0-e^{-(\lambda_2+\lambda_3)t_2}+0=1-e^{-(\lambda_2+\lambda_3)t_2}$$
$$\lim_{t_2 \rightarrow \infty}f(t_1,t_2) = 1-e^{-(\lambda_1+\lambda_3)t_1}-e^{-\infty}+e^{-\infty} = 1-e^{-(\lambda_1+\lambda_3)t_1}-0+0 = 1-e^{-(\lambda_1+\lambda_3)t_1}$$
Is it good like that or is all wrong? :(
 A: It is important in this calculation to make the distinction between the cummulative distribution function (CDF) and the probability density function (PDF). Recall that for a random variable, $X$, in $1$-dimension these are defined respectively as
\begin{align*}
 F(x) &= \mathbf{P}[X \leq x] & \qquad \text{the CDF,} \\
f(x) & = \frac{d}{dx}F(x) & \qquad \text{the PDF.}
\end{align*}
The purpose of working with the PDF is that we often use it to calculate probabilities such as
$$\mathbf{P}[a \leq x \leq B] = \int_a^b f(x) dx.$$
Note that one property of the CDF is that
$$\lim_{x \rightarrow \infty} F(x) = \lim_{x \rightarrow \infty} \mathbf{P}(X \leq x) = 1,$$
to prove this rigorously requires a knowledge of measure theory, but hopefully this is clear heuristically. $F$ is an increasing function, bounded above by $1$,and moreover must approach $1$ as it is a probability measure.
In your question you have swapped between the upper case $F$ and lower case $f$; from what I understand the first equation you give (in $F$) is correct, and you are working with the CDF of a multivariate distribution $(T_1,T_2)$.
In this case we interpret
$$F(t_1, t_2) = \mathbf{P}[T_1 < t_1, T_2 < t_2],$$
and taking the limit in $T_2$
$$\lim_{t_2 \rightarrow \infty} F(t_1,t_2) = \lim_{t_2 \rightarrow \infty} \mathbf{P}[T_1 < t_1, T_2 < t_2] = \mathbf{P}[T_1 < t_1],$$
which is analogous to the heuristic for the limit in $x \rightarrow \infty$ described in the univariate case above.
With this theory in place, your solution above is accurate.
