# Extended limit laws for step by step evaluation of limits

Most textbooks of calculus / real analysis present a version of limit laws which help us to infer about limit of sum or product of functions provided the limits of individual functions are known to exist. Such rules also go by the name algebra of limits.

In this post and my answer to it I provide an extended version of these laws which are more helpful in step by step evaluation of a limit. Before I state those laws it is better to give some introductory remarks.

If $$f$$ is a real valued function defined in certain deleted neighborhood of point $$a$$ then the limiting behavior of $$f(x)$$ as $$x\to a$$ can be of one of the following types:

• $$\lim_{x\to a} f(x)$$ exists. Although it is redundant to state, to avoid confusion / ambiguity this means that the limit exists as a finite real number. We also say that $$f(x)$$ converges to a real number as $$x\to a$$. Example $$\lim_{x\to 0}x$$.
• $$f(x) \to \infty$$ as $$x\to a$$. We say that $$f(x)$$ diverges to $$\infty$$ as $$x\to a$$ and some prefer to write this in symbols as $$\lim_{x\to a} f(x) =\infty$$. Example $$\lim_{x\to 0} (1/x^2)$$.
• $$f(x) \to - \infty$$ as $$x\to a$$. We say that $$f(x)$$ diverges to $$-\infty$$ as $$x\to a$$ and some prefer to write this in symbols as $$\lim_{x\to a} f(x) =-\infty$$. Example $$\lim_{x\to 0} (-1/x^2)$$.
• $$f(x)$$ oscillates finitely as $$x\to a$$. More formally this means that $$f$$ is bounded in some deleted neighborhood of $$a$$ and there exist at least two distinct real numbers $$A$$ and $$B$$ and two sequences $$\{a_n\}, \{b_n\}$$ of numbers in the deleted neighborhood of $$a$$ such that $$\lim_{n\to\infty} a_n=a=\lim_{n\to\infty} b_n$$ and $$\lim_{n\to\infty} f(a_n) =A, \lim_{n\to\infty} f(b_n) =B$$ Example $$\lim_{x\to 0}(1/x)-\lfloor 1/x\rfloor$$.
• $$f(x)$$ oscillates infinitely as $$x\to a$$. This means that there is a sequence $$\{a_n\}$$ of numbers in deleted neighborhood of $$a$$ such that $$\lim_{n\to\infty} a_n=a, \lim_{n\to \infty} |f(a_n) |=\infty$$ and yet neither $$f(x) \to\infty$$ nor $$f(x) \to-\infty$$ as $$x\to a$$. Example $$\lim_{x\to 0}(1/x)\sin(1/x)$$.

The above list is exhaustive and consists of mutually exclusive possibilities. Sometimes the second and third options are combined together and one says that $$f(x)$$ diverges as $$x\to a$$. Similarly the fourth and fifth options can be combined to say that $$f(x)$$ oscillates as $$x\to a$$.

Now we come to extended limit laws.

Theorem 1: Let $$f, g$$ be functions defined in a certain deleted neighborhood of $$a$$ and let $$\lim_{x\to a} f(x)$$ exist and be equal to $$L$$. Then the limiting behavior of $$f(x) \pm g(x)$$ as $$x\to a$$ is of exactly the same type as that of $$g(x)$$ and we can write $$\lim_{x\to a} \{f(x) \pm g(x) \}=\lim_{x\to a} f(x) \pm\lim_{x\to a} g(x) =L\pm\lim_{x\to a} g(x)$$ The case of divergence can be same or opposite (as regards to sign of $$\infty$$) depending on the sign $$\pm$$ which combines $$f, g$$.

Theorem 2: Let $$f, g$$ be defined in a certain deleted neighborhood of $$a$$ and let $$\lim_{x\to a} f(x) =L\neq 0$$. Then the limiting behavior of $$f(x) g(x)$$ as $$x\to a$$ is of exactly the same type as that of $$g(x)$$ and we can write $$\lim_{x\to a} f(x) g(x) =\lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x) =L\lim_{x\to a} g(x)$$ The case of divergence can be same or opposite according as $$L>0$$ or $$L<0$$. Also the case of convergence holds when $$L=0$$ but other cases can't be guaranteed when $$L=0$$.

Both these theorems can be used to evaluate the limit of a complicated expression in a step by step manner by handling one term or one factor at a time whose limit is known thereby reducing the expression to a simpler form at each step. Each step is justified on the basis of the term/factor whose limit is known irrespective of the behavior of other terms/factors.

Moreover the theorems indicate that each step is reversible and hence holds unconditionally. This is better than using the standard limit laws which basically say that the limit has to be applied simultaneously on each part of the expression on the condition that each part has a limit and parts occurring as denominator have non-zero limit.

I will provide proof of one of the theorems as an answer (to be marked community wiki). I expect users to provide other point of views regarding these theorems and any improvements in my question and answer are also welcome.

Note: The above is a more formal and detailed version of the rules presented in this answer and it is based on a request in a comment to another question.

• I use the terms "exists finitely" for case $1$ (or "converges"), "exists infinite" for case $2$ and $3$ (or "diverges") and doesn't exist otherwise. Your definitions are more detailed!
– user
Oct 25, 2018 at 19:08
• I was searching for something I'd written and I came across something ELSE I'd written, and it occurred to me that this other thing I'd written might be of interest to you, at least if you wind up extending your formulations (or use another answer to extend your formulations) to limsup/liminf descriptions. (Presently, I'm not sufficiently interested or have time to do this myself.) See my answer to Sufficient condition for equality of limsups. Nov 7, 2018 at 8:04
• @VerkhovtsevaKatya : also don't generalize a term to a linear combination. Jan 16, 2020 at 1:51
• @VerkhovtsevaKatya: I recommend only one book for those who read my answers (on calculus) and blog: A Course of Pure Mathematics by G H Hardy, 10th edition. Jan 16, 2020 at 10:39
• @Koro: you can also see that $fg$ won't converge because if it did converge to $L$ then $g=fg/f$ converges to $L/\infty=0$ (I hope you can do this rigorously). In short all your claims are correct and do try to give a rigorous proof. If you face any issue you can ask as a separate question. This will also help future readers of this site. Apr 5, 2021 at 9:34

This is a proof of theorem 2 and is presented more as an outline and the details can be easily filled by anyone familiar with proofs of usual limit laws.

The case of convergence is the one usually presented in common textbooks and for this case it can be written explicitly as

Let $$\lim_{x\to a} f(x) = L\neq 0$$. The limit of $$f(x) g(x)$$ as $$x\to a$$ exists if and only the limit of $$g(x)$$ as $$x\to a$$ exists.

The "if" part is a direct application of usual product rule of limits. The "only if" part is also a consequence of the usual algebra of limits (mainly the quotient rule). Just write $$h(x)=f(x) g(x)$$ and then since $$\lim_{x\to a} f(x) =L\neq 0$$ by quotient rule we have $$\lim_{x\to a} g(x) =\lim_{x\to a} \frac{h(x)} {f(x)} =\frac{1}{L}\lim_{x\to a} h(x)$$

If $$L=0$$ then the "if" part holds but not the "only if" and thus the process of splitting the limits is not reversible.

Next we deal with the case of divergence and here we assume $$L>0$$. We then show that $$f(x) g(x)$$ diverges to $$\infty$$ as $$x\to a$$ if and only if $$g(x)$$ does so. First we deal with "if" part and thus let $$g(x) \to \infty$$ as $$x\to a$$. And consider any arbitrary $$M>0$$. By existence of limit of $$f$$ there is a $$\delta_1>0$$ such that $$f(x) >L/2$$ whenever $$0<|x-a|<\delta_1$$ (this is done by choosing $$\epsilon=L/2>0$$ in definition of limit). Again since $$g(x) \to\infty$$ there is a $$\delta_2>0$$ such that $$g(x) > 2M/L$$ (note the $$L\neq 0$$) whenever $$0<|x-a|<\delta_2$$. If $$\delta=\min(\delta_1,\delta_2)$$ then for $$0<|x-a|<\delta$$ we have $$f(x) g(x) >(L/2)(2M/L)=M$$ so that $$f(x) g(x) \to\infty$$ as $$x\to a$$. The case when $$g(x) \to - \infty$$ is similar. For $$L<0$$ the conclusion is same apart from a reversal of sign of infinity.

The "only if" part for divergence can be proved using the "if" part itself. Consider $$G(x) =f(x) g(x), F(x) =1/f(x)$$. Then by quotient rule $$F(x) \to 1/L$$ and $$1/L$$ has same sign as $$L$$. Thus by the proof in last paragraph $$g(x) =F(x) G(x)$$ diverges if $$G(x) =f(x) g(x)$$ diverges.

The case of oscillation presents no significant difficulty. Let's handle the case of finite oscillation here. We show that $$f(x) g(x)$$ oscillates finitely as $$x\to a$$ if and only if $$g(x)$$ does so. We prove the "if" part and the "only if" part can be deduced from the "if" part as done in case of divergence. Since $$f(x) \to L$$ therefore $$f$$ is bounded in some deleted neighborhood of $$a$$. And since $$g(x)$$ oscillates finitely it is also bounded. The product $$f(x) g(x)$$ is thus bounded in some deleted neighborhood of $$a$$. There exist numbers $$A, B, A\neq B$$ and sequences $$a_n, b_n$$ such that $$a_n\to a, b_n\to a$$ and $$g(a_n) \to A, g(b_n) \to B$$. Since $$f(x) \to L$$ we have $$f(a_n) \to L, f(b_n) \to L$$ and by product rule for sequences we have $$f(a_n) g(a_n) \to LA, f(b_n) g(b_n)\to LB$$ Since $$L\neq 0$$ and $$A\neq B$$ we have $$LA\neq LB$$ and thus $$f(x) g(x)$$ also oscillates finitely as $$x\to a$$.

The crucial part to note here is that in each case the condition $$L\neq 0$$ makes the whole process reversible ie it changes "if" (one way implication) to "if and only if" (two way implication). This is important to validate the step by step application of the rule unconditionally.

A proof of theorem 1 is similar and one should try to understand why we don't need the requirement $$L\neq 0$$.