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Given: $$ f(x) = x\left(\frac{(x+1)^3}{(x+2)^3} - 1\right) $$ I need to find a mistake in any of the transitions below (each "$=$" sign has a number above): $$ \lim_{x \to \infty} f(x) \overset{1}{=} \lim_{x \to \infty}x\left(\left(1-\frac{1}{(x+2)}\right)^3 - 1\right) \overset{2}{=} \lim_{x \to \infty} x((1-0)^3-1) \overset{3}{=} \lim_{x \to \infty}x \cdot0 \overset{4}{=} \lim_{x \to \infty}0 \overset{5}{=} 0. $$ I know the real answer is $-3$, so some of those transitions is/are incorrect. My guess is that the 4th transition is incorrect because the $0$ there is not a real zero but something that is very close to $0$. And in such cases we can't multiply infinity by such "zero", that is limit arithmetic is not valid in such cases.

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    $\begingroup$ The problem is at 2. You can't let one $x$ go to infinity without all of the $x$'s going to infinity. You have an indeterminate form here and it'll take a L'hospital trick (or something.) $\endgroup$ – B. Goddard Nov 24 '16 at 13:35
  • $\begingroup$ the error is in equalities 2 and 4. that's an indefinate form. $\endgroup$ – hamam_Abdallah Nov 24 '16 at 13:40
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Your idea is right, but your application isn't. If the $0$ in that limit isn't a "real" zero, then it shouldn't be there - in mathematics, if you say something you don't mean you're saying the wrong thing. You're right that that isn't "really" zero - but that means that an earlier step was wrong. In this case, step 2. $\frac{1}{x + 2}$ does indeed go to zero as $x$ goes to infinity; but it never actually hits zero, so it's incorrect to replace it with zero while still inside the limit.

EDIT: To address your comment below: No, a real number divided by $x$ goes to zero as $x$ goes to infinity. The limit $\lim_{x \to \infty}(2+\frac{1}{x})$ works because the $2$ can be pulled out of the limit to make $2 + \lim_{x \to \infty}\frac{1}{x}$, and then you can use the fact that $\lim_{x \to \infty}\frac{1}{x} = 0$. The issue is that, strictly speaking, it is never correct to replace an expression with its limit without deleting the limit, because the expression need not ever actually take on its limiting value. In many cases, like $\lim_{x \to \infty}(2 + \frac{1}{x})$, breaking this rule is harmless because there isn't any conflict. But when you have what's called an indeterminate form (for example, a product in which one piece goes to infinity and the other one goes to zero) it becomes important that the pieces never take on their values.

The thing is, the statement $\lim_{x \to \infty}x \cdot 0 = 0$ is correct - so it can't possibly be the error. Mathematical statements stand on their own, you don't have to look at the ones before them to see if they're correct. The error is in step $2$, when you replace an expression with its limit, despite the fact that its limit is never obtained.

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  • $\begingroup$ say, if we had an expression $lim_{x \to \infty} (2 + \frac{1}{x})$ wouldn't it be $2$? So we can assume that a real number divided by infinity is zero individually. But the problem I guess is that indeed some parts of the expression with $x$'s become $0$ while part of the $x$'s remain as infinity. So technically wouldn't 4 transition be the problem? $\endgroup$ – Yos Nov 24 '16 at 13:46
  • $\begingroup$ @Yos I've made an edit to my answer responding to your comment; my response was too long to put it in this comment. $\endgroup$ – Reese Nov 24 '16 at 14:20
  • $\begingroup$ Thanks! By the way, why $\lim_{x \to \infty} x \cdot 0 = 0$? Shouldn't it be $\infty$? As far as I know according to infinity arithmetic rules $\infty \cdot c = \infty$ where $c$ is a constant. Isn't $0$ technically a constant? Edit: I just checked the textbook and the arithmetic is only defined for $c$ less than or greater than $0$. So I don't know :) $\endgroup$ – Yos Nov 24 '16 at 14:46
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    $\begingroup$ @Yos You really shouldn't think in terms of using actual $\infty $. The limit is 0 because $x \cdot 0$ is actually equal to zero for all finite $x $. $\endgroup$ – Reese Nov 24 '16 at 14:48

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