# Assuming the existence of solutions in solving exercises

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?

• Yeah the argument is fine as it is, you don’t need epsilon delta. It’s just that the argumeng is formulated in the wrong direction, but in this case it doesn’t matter since its all iff’s. The correct direction is to say that the limit of x-1 is 6, (because x-1 is continous, which at some level you do need to prove by epsilon delta), and then use the theorem that if two functions f and g agree everywhere wxcept at c, then the limit of the two functions as x approaches c is the same. – Ovi Jan 9 '19 at 16:01
• Since a derivative is defined as a limit the existence of derivative is basically an existence of limit. I hope you are referring to this at the end of your question. – Paramanand Singh Jan 9 '19 at 18:53
• $\epsilon - \delta$ definition is used for proving limit theorems and these theorems are then used while evaluating limits. The point of $\epsilon - \delta$ exercises is to remove the psychological fear (if any) of these Greek symbols (by getting familiar with them) and not to learn proof techniques. The proof techniques are better learnt while studying $\epsilon - \delta$ proofs of various theorems. – Paramanand Singh Jan 9 '19 at 18:58

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.)

• To elaborate on this answer: E-mu's original suggested $\lim_{x \to 7} \dfrac{x^2 - 8x + 7}{x - 7} = \lim_{x \to 7} x - 1$, as written, literally does assume the limits exist; but in fact it is widely understood as shorthand for exactly the argument that you elaborate. – LSpice Jan 10 '19 at 2:30
• To the proposer: We can write it in full detail as $\lim_{x\to 7, x\ne 7}(x^2-8x+7)/(x-7)=6\iff \lim_{x\to 7,x\ne 7}(x-1)=6.$ – DanielWainfleet Jan 10 '19 at 17:03

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 $$f(x)$$.

• When I teach this, I call it the “Limits don't see the point” theorem: If $f(x) = g(x)$ for all $x \neq a$, and $\lim_{x\to a}g(x) = L$, then $\lim_{x\to a}f(x) = L$. You can prove this with $\epsilon$-$\delta$. – Matthew Leingang Jan 9 '19 at 17:50
• I like that name ( limits do not see the point theorem ) – Mohammad Riazi-Kermani Jan 9 '19 at 18:14

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 $$x=x_0$$.

• What makes a definition 'safe' (or un-)? – LSpice Jan 10 '19 at 2:30
• I say, for example in the cases that the definition is exactly and explicitly equivalent to the existence or non-existence of a limit not through implicit conditions.... – Mostafa Ayaz Jan 10 '19 at 10:08

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.)

• +1. For beginner Calculus students, I usually even write the last line as "so $\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7}$ exists and is equal to $6$" to make sure that a nontrivial part of the problem was to show that the limit exists. – Taladris Jan 10 '19 at 4:37

Evaluation of a limit in step by fashion based on limit laws also proves that the limit exists (or does not exist depending on final outcome).

The meaning of the first step in your evaluation $$\lim_{x\to 7}\frac{x^2-8x+7}{x-7}=\lim_{x\to 7}x-1$$ is that the LHS of the above equation exists if and only if the RHS exists. To elaborate further the above statement signifies that the limiting behavior of the function $$(x^2-8x+7)/(x-7)$$ as $$x\to 7$$ is exactly the same as that of function $$(x-1)$$ so that if one converges (diverges / oscillates) so does the other. Any algebraic manipulation which is valid for $$x\neq a$$ can be used as a reversible step when evaluating the limit of a function $$f(x)$$ as $$x\to a$$ and your first step is of this kind.

The second step $$\lim_{x\to 7}x-1=6$$ uses standard limits $$\lim_{x\to a} x=a, \lim_{x\to a} k=k$$ and limit rule dealing with difference of functions and in this step the evaluation of limit is complete.

In general when one evaluates the limit of a complicated function in step by step manner then one must ensure that each step (except the last one which removes the limit operator) must be unconditionally true / reversible / of type if and only if. Usually this fact is not stated but rather assumed implicitly. And therefore while performing step by step evaluation of a limit there is no need to prove apriori that the limit in question exists.

Note 1: The use of L'Hospital's Rule is not unconditional / reversible and hence the steps are logically correct only when the method succeeds in giving an answer. If the application of L'Hospital's Rule does not give a definite answer then it does not mean that the original limit does not exist. So when one uses such a technique like L'Hospital's Rule (which works in one direction) then it is better to ensure (via rough work) that it succeeds and then write the step by step evaluation.

Note 2: The usual limit laws (also known as algebra of limits) as stated in most common textbooks are not reversible (you should see their statements which assume the existence of some limits to conclude the existence / evaluation of related limits). These standard formulations can be replaced by their reversible counterparts which are described in this question.