Step 1: State the continuity requirement.
For continuity at $x = \pi$, we need $f(\pi) = \displaystyle\lim_{x \to \pi} f(x)$, where $f(x) = \dfrac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$.
Step 2: Shift the variable.
Put $x = \pi + h$ so that $h \to 0$. Then $\sin x = \sin(\pi + h) = -\sin h$ and $\cos x = \cos(\pi + h) = -\cos h$.
Step 3: Rewrite numerator and denominator.
Numerator $= 1 - (-\sin h) + (-\cos h) = 1 + \sin h - \cos h$. Denominator $= 1 + (-\sin h) + (-\cos h) = 1 - \sin h - \cos h$.
Step 4: Apply half-angle forms.
Using $1 - \cos h = 2\sin^2\tfrac{h}{2}$ and $\sin h = 2\sin\tfrac{h}{2}\cos\tfrac{h}{2}$: numerator $= 2\sin\tfrac{h}{2}\left(\sin\tfrac{h}{2} + \cos\tfrac{h}{2}\right)$, denominator $= -2\sin\tfrac{h}{2}\left(\cos\tfrac{h}{2} - \sin\tfrac{h}{2}\right)$. Here $1 - \sin h - \cos h = (1-\cos h) - \sin h = 2\sin^2\tfrac h2 - 2\sin\tfrac h2\cos\tfrac h2$.
Step 5: Cancel and take the limit.
Cancelling $2\sin\tfrac{h}{2}$, the ratio becomes $\dfrac{\sin\tfrac{h}{2} + \cos\tfrac{h}{2}}{\sin\tfrac{h}{2} - \cos\tfrac{h}{2}}$. Letting $h \to 0$ gives $\dfrac{0 + 1}{0 - 1} = -1$.
Step 6: Conclude the value.
Therefore $f(\pi) = -1$ for the function to be continuous.
\[ \boxed{f(\pi) = -1} \]