Pullback of 2 functors one of which creates limits and the other preserves I have been solving this problem and do not know what to do next and if I am thinking in the right direction.
Suppose we have this pullback square and know that $H$ creates and $G$ preserves limits.
The exercise is to prove that then $H'$ creates limits.
$$
\require{AMScd}
\begin{CD}
A' @>H'>> A \\
@VG'VV @VVGV\\
X' @>H>> X
\end{CD}
$$
Here is my (unfinished) solution:
Let $R: J \to A'$ be a functor such that the composite $H'R$ has a limit in $A$. Now I need to show that $R$ has limit in $A'$. 
Since $G$ preserves limits, the composite $GH'R$ has a limit. But $GH'R = HG'R$, so $HG'R$ also has a limit. Since $H$ creates limits, it means that $G'R$ has a limit. 
Then I look at 2 constant functors $F:J \to A$ and $F':J \to X' $, taking anything to $\lim H'R$ and $\lim G'R$ respectively.
Since this is a pullback diagram, there exists $T:J \to A$ such that $F = H'T$.
That means that the image of $T$ in $A$ is some object $y$ such that $H'y = \lim H'R$.
Now I am trying to prove that this $y$ is in fact $\lim R$.
$T$ acting on morphisms brings cone.
But I do not know how to prove that this cone is universal. I would be very grateful for your help! 
Thank you in advance.
Note: this is an exercise from MacLane, V.6.3
 A: Suppose $D:J\to A'$ is a diagram in $A'$.
Let $(x,\sigma : \Delta_x \to H'D)$ be a limiting cone in $A$.
Since $G$ preserves limits, 
$(Gx,G\sigma : \Delta_{Gx}\to GH'D)$ is a limiting cone for $GH'D=HG'D$.
Since 
$H$ creates limits, there is a unique cone $(y,\tau:\Delta_y\to G'D)$ such that 
$Hy=Gx$ and $H\tau = G\sigma$, and moreover, this cone is a limiting cone for $G'D$.
Then $((x,y),(\sigma,\tau))$ is a cone over $D$ in the fiber product (given by components). Moreover, by uniqueness of $(y,\tau)$, this cone is the unique cone with 
$H'(x,y)=x$ and $H'(\sigma,\tau)=\sigma$.
If we had any other cone (again, given by components) $((x',y'),(\sigma',\tau'))$,
then by definition, $(x',\sigma')$ would be a cone over $H'D$
and $(y',\tau')$ would be a cone over $G'D$ such that $(Gx',G\sigma')=(Hy',H\tau')$ is a 
cone over $GH'D=HG'D$. Therefore there are unique maps 
$a : x'\to x$ and $b:y'\to y$ such that $Ga=Hb$, and $\sigma \circ \Delta_a =\sigma'$,
and $\tau\circ\Delta_b = \tau'$. Thus $(a,b) : (x',y')\to (x,y)$ is a valid map in the fiber product, and has the property that 
$(\sigma,\tau)\circ \Delta_{(a,b)} = (\sigma',\tau')$, and moreover $(a,b)$ is unique with this property.
Thus $((x,y),(\sigma,\tau))$ is a limiting cone, as required.
Some comments on your approach 
You need to be a bit more careful about the uniqueness statement in the definition of creates limits. Also $T$ doesn't act on morphisms in the cone. $T$ is a diagram. Not sure what's going on with that stuff. You may be struggling to finish the proof, since you don't seem to have an explicit description of the morphisms in the limit cone you claim to have constructed in your question.
