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If I know that:

$$ \lim_{x \to \infty } f(x) = 0 $$ and I want to perform: $\lim_{x \to \infty} g(x) = (x+f(x))^3- x^3$

Can I do that: $\lim_{x \to \infty} g(x) = (x+0)^3- x^3 = x^3 - x^3 = 0$ ?

Thanks :)

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  • $\begingroup$ your last line of calculations has an equality that is wrong: you cannot pass to the limit when $\;x\to\infty\;$ in one part and leave $\;x\;$ as it is in other. When this is "done" there must be ajustification proving it can be done. $\endgroup$ – Timbuc Dec 12 '14 at 15:50
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Take $\;f(x)=\frac1x\;$ and get a straightforward counterexample

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    $\begingroup$ You were faster!$~~$ $\endgroup$ – pointer Dec 12 '14 at 15:52
  • $\begingroup$ my calculus is a little rusty, I have an idea as to the answer, can you verify that I am right? $\endgroup$ – Malachi Dec 12 '14 at 16:02
  • $\begingroup$ @Malachi Yes. Will you post a question? $\endgroup$ – Timbuc Dec 12 '14 at 16:03
  • $\begingroup$ It can even converge too a non-zero limit : i.e with $f(x) = \frac{1}{x^2}$ so this is quite much randomness. $\endgroup$ – servabat Dec 12 '14 at 16:04
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    $\begingroup$ I need to look at my Calculus book again....it's been a while @Timbuc. Thank you $\endgroup$ – Malachi Dec 12 '14 at 16:10
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No. Take for example $f(x)=\frac1x$ then

$$(x+f(x))^3-x^3=3x+\frac3x+\frac1{x^3}\xrightarrow{x\to\infty}+\infty$$

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  • $\begingroup$ Hi, can you share how you developed this expression? thanks :) $\endgroup$ – FigureItOut Dec 12 '14 at 16:32
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    $\begingroup$ I used the identity: $$(a+b)^3=a^3+3a^2b+3ab^2+a^3$$ $\endgroup$ – user63181 Dec 12 '14 at 16:34
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$$(a+b)^3-a^3=3 a^2 b+3 a b^2+b^3$$

Then take $f$:

$$ \lim_{x \to \infty } f(x) = 0 $$

And take an $g(x)$:

$$(g(x)+f(x))^3-g^3(x)=3g^2(x)f(x)+3f^2(x)g(x)+f^3(x)$$

From that you can see that:

  • If you have a bounded limit for $g$ then your limit is $0$
  • If $f(x)g^2(x)$ and $f^2(x)g(x)$ tends to $0$ then your limit is $0$

For that you could make a lot of counterexamples in which pass the limit $f(x)$ is not valid.

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