Is something wrong with the calculation below?

$$ \begin{align} \lim_{x \to \infty} \frac{4x^2}{x-2} &= \lim_{x \to \infty} \frac{4x}{1-2/x} \\ & = \frac{(\lim_{x \to \infty} 4x)}{(\lim_{x \to \infty} 1-2/x)} \\ & = \frac{(\lim_{x \to \infty} 4x)}{1} \\ & = \lim_{x \to \infty} 4x \\ & = \infty. \end{align} $$

I ask because if there isn't then the following would seem correct,

$$ \begin{align} \lim_{x \to \infty} \frac{4x^2}{x-2} - 4x &= \left( \lim_{x \to \infty} \frac{4x^2}{x-2} \right) - \left( \lim_{x \to \infty} 4x \right) \\ & = \left( \lim_{x \to \infty} 4x \right) - \left( \lim_{x \to \infty} 4x \right) \\ & = \lim_{x \to \infty} 4x - 4x \\ & = \lim_{x \to \infty} 0 \\ & = 0. \end{align} $$

But it is not, since

$$ \begin{align} \lim_{x \to \infty} \frac{4x^2}{x-2} - 4x &= \lim_{x \to \infty} \frac{4x^2 - 4x(x-2)}{x-2} \\ &= \lim_{x \to \infty} \frac{8x}{x-2} \\ &= \lim_{x \to \infty} \frac{8}{1-2/x} \\ &= 8. \\ \end{align} $$

In what way is this wrong? Where is the mistake?

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    $\begingroup$ I think you have pretty well pinned down the mistake by your careful exposition. After all, $x$ and $x+8$ both tend to infinity, but that doesn't mean that you can find the limit of their difference by taking the difference of the two (infinite) limits. $\endgroup$ – hardmath Nov 25 '16 at 1:20
  • $\begingroup$ But getting to the limit of their difference is exactly how I would find the difference of their limits. I was thinking despite the first calculation being correct, the mistake is substituting something for infinity in the second calculation. $\endgroup$ – henriq Nov 25 '16 at 1:31
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    $\begingroup$ The limit of the difference $(x+8)-x$ is obvious (since the expression simplifies to a constant). But it doesn't make sense to subtract $\infty - \infty$, the difference of the limits. Something that makes sense is not best found by asking for something that does not make sense. $\endgroup$ – hardmath Nov 25 '16 at 1:34
  • $\begingroup$ That makes sense. So can I or can't I use the rules for calculating with limits when one of the terms tends to infinity? And if I can, what is the mistake after all? $\endgroup$ – henriq Nov 25 '16 at 1:44
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    $\begingroup$ @henriq.cd You can do operations with limits unless you have an indeterminate form. In your first calculation you have $\infty/1$ which is not indeterminate ($\infty/1=\infty$). Look at Jack's wikipedia link for the rules and also you may look here for a list of indeterminate forms $\endgroup$ – Momo Nov 25 '16 at 2:13

It is incorrect that

$$ \lim_{x \to \infty} \frac{4x^2}{x-2} - 4x = \left( \lim_{x \to \infty} \frac{4x^2}{x-2} \right) - \left( \lim_{x \to \infty} 4x \right) $$

since in order to apply the limit rule $$ \lim_{x\to\infty}\big(f(x)-g(x)\big)=\lim_{x\to\infty}f(x)-\lim_{x\to\infty}g(x) $$ one needs both at least one of $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}g(x)$ being real numbers. ([Added later:] In the case $\infty-(-\infty)$, one needs to define the arithmatic operation for the extended real numbers first.)

[Added:] It is dangerous to view $\infty$ as a number unless you know exactly what extended real numbers are and what arithmetic operations are (and are not) allowed for them. There are more examples for nonsense by considering $\infty$ as a real number: $$ \lim_{x\to\infty}\left(x\cdot\frac{1}{x}\right)=\infty\cdot 0=0; $$ $$ \lim_{x\to\infty}\big(2x-x\big)=\infty-\infty=0 $$

[Added later:] Also, the step $$ \lim_{x \to \infty} \frac{4x}{1-2/x} = \frac{(\lim_{x \to \infty} 4x)}{(\lim_{x \to \infty} 1-2/x)} $$ would not be mathematically correct unless one is working in the extended real numbers and has defined what is $\dfrac{\infty}{a}$ for a non-zero real number $a$.

  • $\begingroup$ Splitting works even if the limits are infinite, unless you have an indeterminate form. In this case there is an undeterminate form. $\endgroup$ – Momo Nov 25 '16 at 1:18
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    $\begingroup$ Much better after correction. Although "you need at least one of $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}g(x)$ being real numbers." is still not correct. For example $\infty+\infty$ is not undeterminate. $\endgroup$ – Momo Nov 25 '16 at 2:03

You can't split a limit term-by-term like that unless both terms converge. In this case, neither does, so you can't.


$\infty-\infty$ is an indeterminate form, so the result of it can be anything.

Your mistake is exactly here:

$$\left( \lim_{x \to \infty} 4x \right) - \left( \lim_{x \to \infty} 4x \right)= \lim_{x \to \infty} (4x - 4x)$$

On the left side you have $\infty-\infty$, which has to be solved on a case by case basis by returning to the original limit. So splitting in this case does not work.

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    $\begingroup$ No - their mistake is before that, when they split the limit term by term without making sure that the limits of the terms exist. $\endgroup$ – user361424 Nov 25 '16 at 1:15
  • $\begingroup$ It works even if the limits are infinite, unless you have an undeterminate form. And this case has an undeterminate form, as I mentioned. $\endgroup$ – Momo Nov 25 '16 at 1:17

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