TITLE: Assuming the existence of solutions in solving exercises
QUESTION [15 upvotes]: A sizeable chunk of my first calculus course at university comprised of learning techniques to evaluate limits, such as this simple example, evaluating the limit: $$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7}.$$
A typical solution would be to identify that for $x \neq 7$, $$\frac{x^2 -8x + 7}{x-7} = x-1,$$ so $$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7} = \lim_{x \to 7} x-1 = 6.$$
In my eyes, we have shown that if the limit exists, its value must be $6$. We have not shown that the limit exists in the first place and is equal to $6$, since we have presupposed the existence of the limit when writing 

$$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7} = \lim_{x \to 7} x-1,$$

since the existence of both objects on either side of an equality is a necessary condition for the equality to be true (right?).
My main questions are: do such methods of evaluation serve as evidence that these limits in fact exist in the first place, or do they only tell us what the limit ought to be, and the only way we can be sure is to formally prove it using the $\epsilon$-$\delta$ definition? Is this case similar to "finding" the derivatives of functions?

REPLY [3 votes]: You haven't really presupposed the existence of the limit and I would say your argument is fine.  If you want to be really really scrupulous, perhaps you could re-order things in the final line of your argument.
We have
$$\frac{x^2 -8x + 7}{x-7} = x-1$$
for $x\ne7$, and
$$\lim_{x\to7}x-1=6\ ,$$
so
$$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7} =6\ .$$
Writing it this way, you have an immediate justification for every limit you claim.  (Though as I already said, I don't think there is anything seriously wrong with the way you've written it.)