If $$\lim_{x \to 1} \frac{x^2 - ax + b}{x - 1} = 5$$, then a + b is equal to:

We have to evaluate the limit

$$\lim_{x \to 1}\dfrac{x^{2}-ax+b}{x-1}=5.$$

Because the denominator $$x-1$$ approaches $$0$$ as $$x \to 1$$, the limit can stay finite only if the numerator also becomes $$0$$ at $$x=1$$. Therefore we must have

$$1^{2}-a(1)+b=0 \quad\Longrightarrow\quad 1-a+b=0 \quad\Longrightarrow\quad b=a-1.$$

Knowing that $$x^{2}-ax+b$$ vanishes at $$x=1$$, the Factor Theorem tells us that $$x-1$$ is a factor of the quadratic. Hence we can write

$$x^{2}-ax+b=(x-1)(x-c)$$

for some real number $$c$$. Multiplying out the right-hand side gives

$$ (x-1)(x-c)=x^{2}-cx-x+ c=\;x^{2}-(c+1)x+c. $$

Comparing coefficients with the original quadratic $$x^{2}-ax+b,$$ we obtain

$$a=c+1 \quad\text{and}\quad b=c.$$

After canceling the common factor $$x-1,$$ the expression inside the limit simplifies to

$$\dfrac{x^{2}-ax+b}{x-1}=x-c.$$

Now we let $$x \to 1$$:

$$\lim_{x \to 1}(x-c)=1-c.$$

But the given limit equals $$5,$$ so we set

$$1-c=5 \quad\Longrightarrow\quad c=-4.$$

Substituting $$c=-4$$ into the relations $$a=c+1$$ and $$b=c,$$ we get

$$a=-4+1=-3 \quad\text{and}\quad b=-4.$$

Finally,

$$a+b=(-3)+(-4)=-7.$$

Hence, the correct answer is Option D.