Limit laws when not both limits exist

Question

In the calculus textbooks I've come across, the limit laws are given on the condition that both individual limits exist.

Is it safe to weaken that condition by saying that they are valid as long as they do not lead to indeterminate forms, and then abusing notation a bit for determinate forms involving $\infty$?

For example, I would say that: $$ \lim\limits_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x\rightarrow a} f(x)}{\lim\limits_{x\rightarrow a} g(x)} $$ and then say that for this purpose $\frac{\infty}{x} = \infty$ and $\frac{x}{\infty}=0$ where $x\in\mathbb R$.

This seems to work for me, but I know calculus and not analysis. So I'm wondering if there are any pitfalls.

Answer 1

The following answer assumes that a function $f(x)$ defined in a certain deleted neighborhood of $x = a$ has the following five options:

1. $f(x) \to L$ where $L$ is a real number as $x \to a$. This is what I mean when I say that $\lim_{x \to a}f(x)$ exists.
2. $f(x) \to \infty$ as $x \to a$.
3. $f(x) \to -\infty$ as $x \to a$.
4. $f(x)$ oscillates finitely as $x \to a$.
5. $f(x)$ oscillates infinitely as $x \to a$.

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Limit laws related to "algebra of limits" are normally presented in the following fashion:

If $\lim_{x \to a}f(x), \lim_{x \to a}g(x)$ exist then

• $\lim_{x \to a}\{f(x) \pm g(x)\} = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$
• $\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$
• $\lim_{x \to a}\frac{f(x)}{g(x)} = \dfrac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}$ provided that $\lim_{x \to a}g(x) \neq 0$

Out of these the first law can be easily modified so that the existence of only one of the limits $\lim_{x \to a}f(x), \lim_{x \to a}g(x)$ is required. Thus we have the following result:

If $\lim_{x \to a}f(x)$ exists the nature of $\lim_{x \to a}\{f(x) \pm g(x)\}$ is same as the nature of $\lim_{x \to a}g(x)$. This means that

• if $\lim_{x \to a}g(x)$ exists then $\lim_{x \to a}\{f(x) \pm g(x)\}$ also exists and $\lim_{x \to a}\{f(x) \pm g(x)\} = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$
• if $\lim_{x \to a}g(x)$ is $\pm \infty$ then $\lim_{x \to a}\{f(x) \pm g(x)\}$ is also $\pm \infty$.
• if $g(x)$ oscillates finitely (or infinitely) then $f(x) \pm g(x)$ also oscillates finitely (or infinitely).

The laws relating to product and quotient of $f(x), g(x)$ can also be extended in a similar fashion where we require the existence of the limit of only one of the functions $f(x), g(x)$.

However in case of products is it important that the limit of one of the functions (say $f(x)$) must be non-zero and then the behavior of the product is same as the behavior of the other function (namely $g(x)$).

The case of quotients is bit difficult to handle and it is better to think of $f(x)/g(x)$ as $f(x)\times (1/g(x))$ and formulate the rules for $1/g(x)$ and use the laws of products mentioned above. For $1/g(x)$ the idea is simple. If $g(x) \to \pm\infty$ then $1/g(x) \to 0$. If $g(x) \to 0$ and $g(x)$ is positive (or negative) then $1/g(x) \to \infty$ (or $1/g(x) \to -\infty$). If $g(x) \to L \neq 0$ then $1/g(x) \to 1/L$. In other cases $1/g(x)$ oscillates.

Note that these rules suffice to handle all cases except the case of quotients when both the functions tend to $0$ or both of them tend to $\pm\infty$. This is where we need some more work (involving algebraic manipulation, standard limits, L'Hospital's Rule, Taylor series, or Squeeze Theorem).

Thus what you think intuitively is almost correct, however it is better not to abuse notation and start doing arithmetical operations with symbol $\infty$.

Answer 2

No this is false.

$\lim_{x \to 0} \dfrac{2+\sin(\frac{1}{x})}{2+\sin(\frac{1}{x})} = 1$ but $\lim_{x \to 0} ( 2+\sin(\frac{1}{x}) )$ does not even exist, neither in the affinely extended real line nor the projective real line, intuitively because it 'oscillates and never goes anywhere'.

Also $\lim_{x \to 0} \dfrac{x^2}{\sin(x)} = 0$ but it is certainly not $\dfrac{ \lim_{x \to 0} x^2 }{ \lim_{x \to 0} \sin(x) }$ because the latter is meaningless in the first place since both numerator and denominator are zero.