Vtyjkyuk

I have the PDE $u_t = u_xx$ (heat equation).

I am then told that, by writing the equation as $(partial_x + (0)partial_t)^2 u = u_t$, we see that its characteristics would follow the path defined by $dfracdxdt = pm infty$.

I wonder how they came to this conclusion? Please kindly explain. :)

For a second order PDE in two variables $x$ and $t$,
$$
a(x,t) u_xx + 2b(x,t) u_xt + c(x,t) u_tt + dots = 0
,
$$
there is an associated quadratic form at each point $(x,t)$,
$$
Q(h,k) = a(x,t) h^2 + 2b(x,t) hk + c(x,t) k^2
,
$$
and a characteristic curve is a curve whose normal vector $(h,k)$ satisfies $Q(h,k)=0$ at each point of the curve.

For the heat equation we have simply
$$
Q(h,k)=h^2
$$
so the normal vector should be $(h,k)=(0,1)$ at each point, which forces the characteristics to be lines of constant $t$.

The rewriting of the PDE collects the leading order (second order) derivatives on the right-hand side, as the characteristics should be determined by the leading order terms.
If you have a vector $v=(a,b)$ in the $(x,t)$-plane, then the characteristics of the second order operator $(vcdotnabla)^2=(apartial_x+bpartial_t)^2$ are lines in the direction of the vector $v$.
In this particular case $a=1$ and $b=0$, so the line points in the $x$ direction with no $t$ component.
This conclusion is then expressed — in my opinion clumsily — by saying that $fracdxdt=pminfty$.

Since you gave no source in your question, I can't interpret everything.
But there is truth to the statement: the heat equation (not only in $1+1$ dimensions) has infinite speed of propagation, meaning that solutions travel infinitely fast in all directions.
This is described in the question as $fracdxdt=pminfty$, but I do hope the source you are using also gives a description in words.