Solving a differential equation, second pair of eyes needed. I have a simple ODE which contains a constant k.  Solving this ODE gives a solution containing k.  If I set $k=0$ in the solution, I do not get the solution that I get if I set $k=0$ in the original ODE.  I have been through my workings over and over, but can't see the error.  A second pair of eyes please!  
Here it is:
$\frac{dT}{dx} = k(T-T_{amb})$ where $T_{amb}=Ax+B$
The boundary condition is:
$T=T_0$ when $x=0$, we need to find $T=T_1$ at $x=L$
Solution for $k\ne0$:
$\frac{dT}{dx} = k(T-(Ax+B)) = kT-kAx-kB$
let $a=k, b=-kA, c=-kB$
So we have to solve: $\frac{dT}{dx} = aT + bx + c$
Solution:
let $v=aT+bx$, so $\frac {dv}{dx}=a\frac {dT}{dx}+b => \frac {dT}{dx}=\frac 1a(\frac {dv}{dx}-b)$
the ODE then becomes: $\frac 1a(\frac {dv}{dx}-b)=v+c$
$=>\frac 1a\frac {dv}{dx}-\frac ba=v+c$
$=>\frac 1a\frac {dv}{dx}=\frac ba+v+c$
$=>\frac {dv}{dx}=a(\frac ba+v+c)$
variable separable, so
$\int_{v_0}^{v_1}\frac{dv}{\frac ba+v+c}=\int_0^L a.dx$
$=>\ln(\frac{\frac ba+v_1+c}{\frac ba+v_0+c})=aL$
back substitute T:
$=>\ln(\frac{\frac ba+aT_1+bL+c}{\frac ba+aT_0+c})=aL$
$=>\frac ba+aT_1+bL+c=e^{aL}(\frac ba+aT_0+c)$
$=>T_1=\frac 1a(e^{aL}(\frac ba+aT_0+c)-\frac ba-bL-c)$
substituting A, B:
$T_1=\frac 1ke^{kL}(-A+kT_0-kB)+A/k+AL+B$
Now for the solution where $k\to0$, one gets, as $e^{kL}\to1$:
$T_1=T_0+AL$
However the solution to the original ODE with $k=0$ gives $\frac{dT}{dx} = 0$
$=>\int_{T_0}^{T_1}dT=0$
$=>T_1=T_0$
Why do I get an extra term $AL$ doing it the first way?
 A: You evaluation of the limit is wrong. By using the first order approximation $e^{kL}\approx1+kL$, the expression simplifies to $T_1=T_0+k(T_0-B)L$ and of course in the limit, $T_1=T_0$.
A: If $k=0$ then $T'=0\iff T=cst$
If $k\neq 0$ then homegeneous equation is $T'=kT\iff T=Ce^{kx}$
Constant variation to find a particular solution with RHS :
$\require{cancel}T'=\cancel{Cke^{kx}}+C'e^{kx}=k(T-T_0)=\cancel{kCe^{kx}}-k(Ax+B)\iff C'=-k(Ax+B)e^{-kx}$
$C=(Ax+B+\frac Ak)e^{-kx}+D$

$T=Ax+B+\frac Ak+De^{kx}$

I think you complicate unnecessarily using your method.

Edit: after problem fix and comment
The solution for $k=0$ is still valid, we get $T=T_0$ and this forces $T_1=T_0$.
When $k\neq 0$ we have to work initial conditions :
We have 
$\begin{cases}
T_0=B+\frac Ak+D \\
T_1=AL+B+\frac Ak+De^{kL}=T_0+AL+D(e^{kL}-1) \\
\end{cases}$
Since $A,B$ are constants independent of $k$, the limit condition $\lim\limits_{k=0}T_1=T_0$ implies that $D$ actually depends on $k$.
$T_1=T_0+AL+D\underbrace{(1+kL+o(k)-1)}_{\text{Taylor Expansion of exponetial}}=T_0+\underbrace{(AL+kDL)}_{\text{need to be }0}+\underbrace{o(kD)}_{\text{need to }\to 0}$
So we need $D=-\frac Ak$.
Substituting in $T$ we get :

$T=T_0-\frac Ak\left(e^{kx}-1-kx\right)$

You can now check that this solution is continuous in $k=0$.
A: You are ignoring some division-by-zero problems in your limit. While $e^{kL}$ indeed tends to $1$, $1/k$ tends to infinity. The rules for limit arithmetic only apply if the limits of the operands actually exist, the difference between two "non-identical" infinities can be anything, that is, it is not well-defined. Assemble like expressions in your solution to get 
$$
T_1=T_0e^{kL}+A\left(L-\frac {e^{kL}-1}k\right)+B\left(1-e^{kL}\right).
$$
Then the  $A$ term is the difference between derivative and difference quotient which indeed goes to zero for $k\to 0$.
