Consider $$\lim_{x\to\infty}\frac{f(x)}{g(x)} + \lim_{x\to\infty}\frac{h(x)}{i(x)}$$ I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as $$\lim_{x\to\infty}\frac{f(x)}{g(x)} +\frac{h(x)}{i(x)}$$ Why is this incorrect?

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    $\begingroup$ Limit operator is not a linear transformation, that is why. $\endgroup$ – Bertrand Wittgenstein's Ghost Nov 29 '18 at 0:20
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    $\begingroup$ I think it only goes wrong when you will end up with something of the form $+\infty - \infty$. So, if you avoid these cases you can apply it if I am correct. $\endgroup$ – Stan Tendijck Nov 29 '18 at 0:54

I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...

This has nothing to do with the L'Hôpital's rule itself.

The rule that you cannot use is:


And you can see from JDMan4444's answer that there are situations where this rule does not work.

However, if you are sure that $\lim\limits_{x\to...}f(x)$ and $\lim\limits_{x\to...}g(x)$ exist (and are not $\pm\infty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)

This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.

  • $\begingroup$ The OP is related to THAT and the main doubt is on the application of HR. $\endgroup$ – user Nov 29 '18 at 7:42

Consider the following example:

$$ \lim_{x\rightarrow\infty}\frac{x^2}{x}+\lim_{x\rightarrow\infty}\frac{-x^2}{x} = \infty - \infty \quad \text{(which is undefined)} $$ $$ \lim_{x\rightarrow\infty}\frac{x^2}{x}+\frac{-x^2}{x} = 0 $$

  • $\begingroup$ Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP. $\endgroup$ – user Nov 29 '18 at 7:44
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    $\begingroup$ Your second line "$\lim_{x\rightarrow\infty}\frac{x^2}{x}+\frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention. $\endgroup$ – user21820 Nov 29 '18 at 13:51
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    $\begingroup$ @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there. $\endgroup$ – user3445853 Nov 29 '18 at 15:17
  • $\begingroup$ @user21820 It could be convention but since the expression $\frac{x^2}{x}+\frac{-x^2}{x}$ is identically equal to zero for all $x\neq 0$ I think that it is a noce assumption read that as $$\lim_{x\rightarrow\infty} \left(\frac{x^2}{x}+\frac{-x^2}{x}\right) =\lim_{x\rightarrow\infty} 0= 0$$ $\endgroup$ – user Nov 29 '18 at 15:29
  • $\begingroup$ @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here. $\endgroup$ – user21820 Nov 29 '18 at 16:47

The following identity

$$\lim_{x\to \infty}\left(\frac{f(x)}{g(x)} +\frac{h(x)}{i(x)}\right)=\lim_{x\to \infty}\frac{f(x)}{g(x)} + \lim_{x\to \infty}\frac{h(x)}{i(x)}$$

doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form

$$\lim_{x\to \infty}\left(\frac{f(x)}{g(x)} +\frac{h(x)}{i(x)}\right)=\lim_{x\to \infty}\frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$

the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.


It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $\infty - \infty$ or $-\infty + \infty$ cases are problematic whether you're contemplating L'Hopital or not.

  • $\begingroup$ The advice was given HERE for that specific example and it was a completely different case to the one you are referring to. $\endgroup$ – user Nov 29 '18 at 7:41
  • $\begingroup$ @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question? $\endgroup$ – zhw. Nov 29 '18 at 19:19
  • $\begingroup$ My aim was solely give you a better context for the question posed. $\endgroup$ – user Nov 29 '18 at 19:23

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