# Simple question regarding limits

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 :)

• 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. – Timbuc Dec 12 '14 at 15:50

Take $\;f(x)=\frac1x\;$ and get a straightforward counterexample

• You were faster!$~~$ – pointer Dec 12 '14 at 15:52
• my calculus is a little rusty, I have an idea as to the answer, can you verify that I am right? – Malachi Dec 12 '14 at 16:02
• @Malachi Yes. Will you post a question? – Timbuc Dec 12 '14 at 16:03
• It can even converge too a non-zero limit : i.e with $f(x) = \frac{1}{x^2}$ so this is quite much randomness. – servabat Dec 12 '14 at 16:04
• I need to look at my Calculus book again....it's been a while @Timbuc. Thank you – Malachi Dec 12 '14 at 16:10

No. Take for example $f(x)=\frac1x$ then

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

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

$$(a+b)^3-a^3=3 a^2 b+3 a b^2+b^3$$

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.