# Is the way we “compute limits” rigorous?

This is a general process question. I'm taking my first analysis class in in the chapters of differentiation and continuity, there were quite a few exercises assigned that essentially come down to "compute such-and-such limit". Now, the definition of a limit is clear but for practical purposes isn't constructive; when I look up answers online, or my professor publishes the suggested answers, invariably what is done is the following:

1) Use some computational trick to reduce the expression whose limit we want to a polynomial or something trivial (factoring, multiplying by some sort of conjugate expression to eliminate radicals, etc). Alternatively, use L'Hospitals rule.

2) Heuristically assume that $\lim_{t \rightarrow x} t^a = x^a$ (heuristic because we justify it in my experience by saying "polynomials are continuous, so the limit must be thus", though in a former thread a member suggested an excellent proof using logarithms:

That being said, this doesn't prove that this is the limit. So my questions are:

1) Are these tools reliable; that is, do they produce limits in general that can be rigorously proven to be correct (using $\epsilon, \delta$ proofs).

2) When is it safe in a proof to simply say: "let such-and-such tend to zero", or "Let this value go to infinity"?

I have some intuitive ideas, but I can't seem to codify any sort of solid answer, and I'm hoping someone with a lot more experience and a bird's eye view on the problem (both in terms of mathematical savvy and teaching experience) can shed some light.

Thanks.

• Saying that the $\lim\limits_{t \rightarrow x} f(t) = f(x)$ when $f$ is continuous is perfectly rigorous, since that is the definition of continuity at $x$. One can also rigorously prove continuity of polynomials. The epsilon-delta arguments can be used to prove trivially that the constants and the function $t$ are continuous, and limit theorems can be used to prove products are also continuous. The continuity of polynomials then follows by induction. – Tob Ernack May 4 '17 at 17:12
• In any decent course in analysis, you prove everything. Everything is done completely rigorously. That doesn't mean you always go back to the definition in proving a limit exists, or equals $1$, or whatever. That's what theorems are for. – zhw. May 4 '17 at 17:20
• You should already proven when you first defined limits, that these operations are all valid operations. – Doug M May 4 '17 at 17:29
• @zhw I'm taking analysis at an Ivy League university, so I think we can assume the course is "decent". That said, the basic fact that for any $a \in \mathbb{R} \lim_{t \rightarrow x} t^a = x^a$ isn't something I've ever seen proven directly except on this site when I asked. Of course we can say 1) All polynomials are continuous 2) therefore this limit as written, and I have proven the continuity of polynomials. – BenL May 4 '17 at 17:35
• I appreciate your honesty both in accepting your beliefs about limits and mention of the flagging. You may wish to have a look at the proofs of various limit theorems which are used to evaluate limits. This will be a first step towards understanding that these things are rigorous. Next you will need a rigorous theory of exponential and logarithmic functions which are almost never presented rigorously in introductory calculus texts. You may find nice exposition of these topics in my blog posts paramanands.blogspot.com/2014/05/… – Paramanand Singh May 4 '17 at 18:57

## 2 Answers

The method is rigorous. It is based on the following theorems

$$\lim\limits_{x\to a}(f(x)+g(x))=\lim\limits_{x\to a}f(x)+ \lim\limits_{x\to a}g(x)$$ $$\lim\limits_{x\to a}(f(x)g(x))=(\lim\limits_{x\to a}f(x)) (\lim\limits_{x\to a}g(x))$$

$$\lim\limits_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to a}f(x)}{ \lim\limits_{x\to a}g(x)}$$

Assuming that $\lim\limits_{x\to a}f(x)$ and $\lim\limits_{x\to a}g(x)$ exist and are finite (and $\lim\limits_{x\to a}g(x)\neq 0$ in the last case).

The aim of the tricks you mention is to express a the function in a form where application of the above theorems is valid.

There is of course input of basic limits (eg $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$) which must be proved by some other means.

• I believe you meant to write $\lim_{x \to 0} \sin(x)/x = 1$ – Alex Provost May 4 '17 at 17:36
• Don't we also have to assume that the limit on the LHS exists as well? – Brevan Ellefsen May 4 '17 at 17:38
• @BrevanEllefsen No that is the conclusion of the theorem. – Rene Schipperus May 4 '17 at 17:43

And there are a number of ways to prove that polynomials are continuous.

One way is to prove that the sum and product of continuous functions are continuous and then prove that 1 and x are continuous.

Another, specialized for polynomials, is to use $x^n-a^n =(x-a)\sum_{k=0}^{n-1} x^k a^{n-1-k}$ so

$\begin{array}\\ |x^n-a^n| &=|x-a||\sum_{k=0}^{n-1} x^k a^{n-1-k}|\\ &\le|x-a|n\max(|x|, |a|)^{n-1}\\ &\le|x-a|n(|a|+1)^{n-1} \qquad\text{if }|x-a| \le 1\\ \end{array}$