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

The solution which I have says that  $x-1$ tends to $0$. Therefore numerator $x^2 - ax + b$ must tend to $0$ because the limit exists.
My query is: why it is written, that limit exists so numerator of tend to $0$.

$$\frac{x^2-ax+b}{x-1}=x-a+1+\frac{b-a+1}{x-1}$$ and for the limit to exist, the numerator of the fraction must be zero.
Hence
$$\begin{cases}b-a+1=0,\\1-a+1=5.\end{cases}$$
