I'm starting to think you can always apply them, but you might get indeterminate form 0/0. However, I haven't seen this stated in relation to limits, only in some complex derivatives. It's not stated when limit properties are applied.

I had expected that you could apply properties and rules blindly (with any preconditions included as part of the property/rule), but maybe you must also look at the outcome you get in order to know what rules you're allowed to apply? Perhaps it's like division in algebra, where a manipulation might be valid (at "compile time"), but if specific values of variables make a divisor equal zero, then it's undefined (at "runtime")?

I'll give a warmup example, then the one I'm actually concerned about.

1) First example: applying the quotient limit property to $x/x$ as $x\to0$, which should be $1$:

$$ \lim_{x \to 0}\frac{x}{x} = \frac{\lim_{x \to 0}x}{\lim_{x \to 0}x} = \frac{0}{0} $$

2) Second example: in part of a proof of the product rule (used by Khan at 7:30, and Spivak in Ch.10, Theorem 4, p.45):

$$ \lim_{h\to0} \frac { f(x+h) [ g(x+h) - g(x) ] } {h} = \lim_{h\to0} f(x+h) \lim_{h\to0} \frac { g(x+h) - g(x) } {h} $$

Then using $\lim_{h\to0} f(x+h) = f(x)$:

$$ = f(x) \lim_{h\to0} \frac { g(x+h) - g(x) } {h} $$

My difficulty is you could apply the same approach to the other factor: $\lim_{h\to0} g(x+h) = g(x)$.

$$ \lim_{h\to0} \frac {f(x+h)} {h} \lim_{h\to0} [ g(x+h) - g(x) ] = \\ \lim_{h\to0} \frac {f(x+h)} {h} [ \lim_{h\to0} g(x+h) - \lim_{h\to0} g(x) ] = \\ \lim_{h\to0} \frac {f(x+h)} {h} [ g(x) - g(x) ] = \\ \lim_{h\to0} \frac {f(x+h)} {h} 0 = 0 $$

Am I wrong to think you should be allowed to apply property limits whenever you like, without condition? (And if there are pre-conditions, they should be part of the property?) Or, is there some aspect relating to $0/0$ (indeterminate form) whose connection to property limits I've somehow missed or not appreciated?

It seems derivatives are about finding a ratio $a/b$, even as $a,b\to0$. But I think limits are meant to be standalone, independent of derivatives.

PS. if my confusion is too fundamental to address in an answer, could you suggest a reference that definitely does address it, please? (It's a lot of work to go through a reference, only to find it doesn't have the answer I need.) Many thanks!

  • 1
    $\begingroup$ Whenever a textbook (or any other source) mentions the limit properties / laws, it is supposed to specify the conditions under which these can be applied. If there is no such mention of the conditions then this is highly irresponsible on their part. On the other hand if the condition are mentioned but someone overlooks them then it is sheer carelessness on the part of the reader. The second phenomenon is more common than the the first one. $\endgroup$ – Paramanand Singh Dec 6 '17 at 8:31
  • $\begingroup$ Unfortunately, the khan video on limit properties for the quotient limit property (at 3:30) doesn't mention that the divisor limit $\neq0$. However, I must admit that even when conditions are mentioned (e.g. that $f$ is differentiable), I don't really pay attention... and the significance of those conditions are not mentioned. The conditions are just stated. $\endgroup$ – hyperpallium Dec 7 '17 at 5:24
  • $\begingroup$ Well you should pay attention to the conditions under which a system works. Otherwise you may get in trouble. Also this advice should be given in textbooks with more emphasis and with examples where the rules fail if conditions are not met. $\endgroup$ – Paramanand Singh Dec 7 '17 at 5:47
  • $\begingroup$ You should also put a comment on khan academy website for this so that they may edit their video for benefit for future viewers. In mathematics even a slight error can cause great problems and people should try not to omit essential details while presenting mathematical content (this applies to content on this website also). $\endgroup$ – Paramanand Singh Dec 7 '17 at 5:53
  • $\begingroup$ Maybe an appropriate way to address this would be exercises, with complex nested limit expressions, but of very simple functions, that require some trial-and-error/backtracking... and not just the known proofs "on rails" - I feel sure this must exist in some textbook, but don't know where. I agree about Khan, but they don't allow comments without an account. Offtopic, but they could allow anon comments with CAPTCHA, if they were more interested in corrections than amassing accounts. $\endgroup$ – hyperpallium Dec 7 '17 at 7:02

Apart from my comment to the question here is an answer specific to the scenario mentioned in your question about product rule of limits.

Let me specify the product rule first (I haven't seen this version mentioned in any textbook, but this is the way it is used in practice):

Extended Product Rule of Limits: Let $f, g$ be real valued functions defined in a certain deleted neighborhood of point $a$ and let $\lim_{x\to a} f(x) $ exist and be non-zero then $$\lim_{x\to a} f(x) g(x) =\lim_{x\to a} f(x) \cdot\lim_{x\to a} g(x) \tag{1}$$ irrespective of the fact whether $\lim_{x\to a} g(x) $ exists or not.

Also the equation $(1)$ has to be interpreted in a more general sense when the $\lim_{x\to a} g(x) $ does not exist. In this case the limit $\lim_{x\to a} f(x) g(x) $ does not exist in the same manner (diverging to infinity or oscillation) as $\lim_{x\to a} g(x) $. The usual product rule as given in most textbooks says that the equation $(1)$ holds if both the limits $\lim_{x\to a} f(x), \lim_{x\to a} g(x) $ exist and the above reformulation relaxes this condition to the existence of limit of one of the functions.

Now you can see why the factor $g(x+h) - g(x) $ can not be separated with the rest of the expression. This factor tends to $0$ and then you can't make use of the product rule irrespective of the limit of the remaining factor. You will have to deal with the part $f(x+h) /h$ also. On the other hand if we take $f(x+h) $ as a factor this can be separated out (if $f(x) \neq 0$) without worrying about the second part $(g(x+h) - g(x)) /h$. If $f(x) =0$ then we need to check the other part also (and here we are given that the other part tends to $g'(x) $ so that the usual product rule given in most textbooks applies).

In general when we are evaluating the limit of a complicated expression we proceed in a step by step manner first separating the terms (this requires sum / difference rule of limits) and then separating the factors of each term. At each step one should be able to make the decision of separating a term or a factor from the rest of the expression by just analyzing the limit of the term/factor under consideration and without worrying about the rest of the expression. In case of a term the only requirement is that the limit of the term under consideration exists but in case of a factor the requirement is that the limit of the factor under consideration must be non-zero. If a factor does tend to zero we can't proceed without analyzing the remaining expression. This is typically how most textbooks present the examples of limit evaluation, but the details are not explained properly.


You only can apply if limits exist with other limitation case by case:


  • $\begingroup$ Oh, so the quotient limit property actually does have a pre-condition forbidding a divisor evaluating to zero. Let me expand my second example. $\endgroup$ – hyperpallium Dec 6 '17 at 7:09
  • $\begingroup$ Yes, in this case the application of algebric rule is forbidden! To prove that you have to consider the definition. $\endgroup$ – gimusi Dec 6 '17 at 7:12
  • $\begingroup$ Could you have a look at my second example? It gets an incorrect result of zero - without using the division rule. $\endgroup$ – hyperpallium Dec 6 '17 at 7:30

You can apply the limit laws like this $$ \lim_{h\to0}\frac{f(x+h)[g(x+h)-g(x)]}{h}=\lim_{h\to0}\frac{f(x+h)}h\lim_{x\to0}[g(x+h)-g(x)] $$ provided that both the limits on the right side above exist and are finite. The problem is, must of the time, that's not going to apply to $\lim_{h\to0}\frac{f(x+h)}h$, but it always applies to $\lim_{h\to0}\frac{g(x+h)-g(x)}h$ (remember that when deriving the product rule we assume $f$ and $g$ are differentiable). This is the reason we split the limit the way we do, and not your way.

  • $\begingroup$ Thanks, as I expanded that example, I was starting to think that if $f(x)\neq 0$, then $\lim_{h\to0} \frac{f(x+h)}{h}$ would end up being infinite... (BTW if $f(x)=0$, you get $f'(x)$ ) So, you do need to look ahead to outcomes (e.g. "is the resulting limit infinite?") which is a post-condition, not a pre-condition. Therefore, you can't just apply the rules blindly, even when including all pre-conditions? $\endgroup$ – hyperpallium Dec 6 '17 at 8:05
  • $\begingroup$ @hyperpallium More or less, yes. What it ends up meaning in practice is that you apply the limit rules, and keep an eye out for whether anything goes wrong. If it turns out it doesn't work, you backtrack, and no damage is done. $\endgroup$ – Arthur Dec 6 '17 at 8:17
  • $\begingroup$ @hyperpallium It's not about "post conditions" and "pre conditions". $\lim (f/g)=\lim f/\lim g$ when $\lim f$ is finite and $\lim g$ is non zero, but that doesn't you have to write down the (wrong) limit of quotients formula and then backtrack. You can separately compute $\lim f$ and $\lim g$ before you look at their quotient at all. $\endgroup$ – Jack M Dec 6 '17 at 9:13
  • $\begingroup$ Thanks! If the operand expressions are themselves complex, it might be some work to determine their value; but I see that with each operator "opened up", the "children" operands are necessarily simpler expressions than their "parent" operator. However... when there are several different ways to factor (perhaps nested), and you have to investigate each one, the backtracking can be complex. BTW overcoming different factorings also occurs in Polynomial Identity Testing $\endgroup$ – hyperpallium Dec 7 '17 at 5:21

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