Assuming the existence of solutions in solving exercises

There's no logical problem with this argument. The expressions $$ \frac{x^2 -8x + 7}{x-7} \text{ and } x-1 $$ are equal when $x \ne 7$, so the first expression has a limit at $7$ if and only if the second does. There is no need to assume the existence of the limit in advance.

Whether or not you need the $\epsilon - \delta$ argument to find the the limit of $x-1$ depends on the level of rigor your instructor requires.

(There are other situations where a correct argument does have the form

the limit is such and such provided the limit exists

usually followed by a separate proof that there is a limit.)

Note that when we find the limit of a function at a point the value of the function at that point is not important.

In your example the function $f(x) =\frac {x^2-8x+7}{x-7}$ and $g(x)=x-1$ have the same values at every point except at $x=7$ thus they have the same limits at that point.

Since $g(x)$ has a limit of $6$ at $x=7$ so does $f(x)$.

The $\epsilon-\delta$ approach is the safest and most standard definition of limit. To show that $\lim_{x\to x_0}f(x)$ exists and is equal to $L$, we need to show that $$0<|x-x_0|<\delta\to |f(x)-l|<\epsilon$$here we need to show that $$0<|x-7|<\delta\to \left|{x^2-8x+7\over x-7}-6\right|<\epsilon$$also note that for $|x-7|>0$ we have $x\ne 7$ therefore $$\left|{x^2-8x+7\over x-7}-6\right|<\epsilon\iff |x-1-6|<\epsilon $$which means that choosing $\delta=\epsilon>0$ we have proved the existence of the limit i.e.$$0<|x-7|<\epsilon\to \left|{x^2-8x+7\over x-7}-6\right|<\epsilon$$therefore$$\lim_{x\to 7}{x^2-8x+7\over x-7}=6$$Comment

You can exploit this definition whenever you wanted to find the limit of $${(x-x_0)g(x)\over x-x_0}$$ in $x=x_0$.