Compute
\[\lim_{x\to +\infty} x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\]

- watchmath

Compute
\[\lim_{x\to +\infty} x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\]

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- anonymous

even with the "x" out front?

- watchmath

yes, the problem is correct

- anonymous

\[\lim_{x\to +\infty} x\times \frac{x}{x}\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\]
\[ x^2 ( \sqrt{1 + \frac{1}{x^2}} + \sqrt[3]{1 + \frac{1}{x^3}})\]
I think it must be \(\infty\)

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## More answers

- anonymous

oh i meant with the x out front it seemed unlikely to be 0

- anonymous

ishaan it is a minus sign. with a plus you would get infinity for sure

- anonymous

it is minus

- TuringTest

lost a minus sign mertsj, right\[x^2 ( \sqrt{1 + \frac{1}{x^2}} - \sqrt[3]{1 + \frac{1}{x^3}})=\infty(0)\]so it is still indeterminate

- TuringTest

oh, ishaan, whoever it was lol

- anonymous

Oh yeah, sorry.
thanks satellite

- anonymous

i will make a guess at a method, but it looks like a lot of work (algebra)
i bet the usual gimmick will work of multiplying by the conjugate, but since you have powers of 1/2 and 1/3 you are going to have to use
\[(a^6-b^6)=(a-b) (a+b) (a^2-a b+b^2) (a^2+a b+b^2)\]

- anonymous

that is just a guess though

- anonymous

*Bookmark*

- anonymous

too much algebra for me, i can't do it

- Mertsj

You can. You just don't want to.

- myininaya

i think i might have it

- watchmath

hello myininaya. Still remember me? :)

- myininaya

yes i do
i'm getting the limit is infinity is that right?

- myininaya

so i will show you what i did...

- anonymous

@watchmath myininaya and i were just mentioning you the other day. sunday

- anonymous

i said i missed your problems (not that i can do them) and no the answer is not infinity, but i cheated.

- watchmath

no, the limit is finite

- anonymous

i think the method i wrote would work if i could do the algebra

- myininaya

\[\lim_{x \rightarrow \infty}\frac{x((x^2+1)^\frac{1}{2}-(x^3+1))^\frac{1}{3}}{1} \cdot \frac{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\]
\[\lim_{x \rightarrow \infty}\frac{x(x^2+1-(x^3+1)^\frac{2}{3})}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\]
\[\lim_{x \rightarrow \infty}\frac{x^3+x-x(x^3+1)^\frac{2}{3}}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\]
So if I divide both top and bottom by x I get
\[\lim_{x \rightarrow \infty}\frac{x^2+1-(x^3+1)^\frac{2}{3}}{(\frac{x^2+1}{x^2})^\frac{1}{2}+(\frac{x^3+1}{x^3})^\frac{1}{3}}\]
\[\lim_{x \rightarrow \infty}\frac{x^2+1-(x^3+1)^\frac{2}{3}}{(1+\frac{1}{x^2})^\frac{1}{2}+(1+\frac{1}{x^3})^\frac{1}{3}}\]
so we have that the bottom goes to 2 as x goes to positive infinity
but the top goes to infinity
so that is why i conclude the answer was infinity
Maybe I should play with the top some more
and hey how are you watchman lol
i have missed your problems

- anonymous

@myininaya your numerator is wrong

- anonymous

middle terms don't add to zero because the powers are not the same. i think it the same gimmick will work, multiply something to clear the radicals, but since one power is 1/2 and the other is 1/3 i am pretty sure you have to use
\[(a^6-b^6)=(a-b) (a+b) (a^2-a b+b^2) (a^2+a b+b^2)\]to clear them

- myininaya

satellite i disagree i think my numerator is right

- anonymous

oh i am sorry, yes it is right, but you still can't get the limit that way

- anonymous

Here's a solution. Using the extension of the binomial theorem to arbitrary complex numbers, you have that \[(1+x^2)^\frac12=x+\frac12x^{-1}-\frac18x^{-3}+O(x^{-5})\] \[(1+x^3)^\frac13=x+\frac13x^{-2}+O(x^{-5})\] Thus, you have that \[\lim_{x\to \infty}x\left((1+x^2)^\frac12-(1+x^3)^\frac13\right)\] \[=\lim_{x\to\infty}x\left(\left(x+\frac12x^{-1}+O(x^{-3})\right)-\left(x+\frac13x^{-2}+O(x^{-5})\right)\right)\] \[=\lim_{x\to\infty}x\left(\frac12x^{-1}+O(x^{-2})\right)=\lim_{x\to\infty}\frac12+O(x^{-1})=\frac12\]

- myininaya

imperialist is a showoff

- anonymous

pretty damned nice

- watchmath

imperialist is correct. Any elementary solution?

- anonymous

I'm pretty sure satellite's method is also a good way to approach it, but as myininaya can attest, the algebra is not very friendly to say the least. Good job though myininaya!

- anonymous

well i am still thinking of writing the difference of two sixth powers, the algebra is killing me

- anonymous

i would say the above approach is much nicer, especially since i have filled up a piece of paper and probably have at least one mistake

- Mertsj

Why don't you brainiacs go help pre-algebra? He's been trying to get help most of the day.

- myininaya

you know numerically my top approaches 1

- myininaya

algebraically my bottom approaches 2

- myininaya

if some how i can show without a numerical approach or graphical approach that the top approaches 1, then I would be dandy

- anonymous

where is the pre-algebra question? i can't find it

- Mertsj

Updated 41 minutes ago.

- myininaya

So how can I show that
\[\lim_{x \rightarrow \infty}(x^2+1-(x^3+1)^\frac{2}{3})=1?\]

- anonymous

Hmm x^2 + 1 - (x^3+1)^(2/3) can't we ignore 1?

- myininaya

couldn't i say \[x^3+1 \approx x^3\]

- myininaya

\[(x^3)^\frac{2}{3}=x^2\]

- myininaya

x^2-x^2=0
we are left with 1

- anonymous

Mhm, sounds good to me!

- myininaya

i think it sounds awesome! :)

- watchmath

If you can compute
\[\lim_{x\to \infty} x\left(\sqrt[3]{x^3+1}-x\right)\]
then you can compute the original problem :D.

- JamesJ

(For what it's worth, I think Imperialist has it. I can't see a proof that doesn't rely on some version of a power series expansion. I've learnt the hard way never to say never, so perhaps there something else more elementary out there. Nonetheless, sometimes elementary problems have complicated solutions.)

- myininaya

James what do you think about what I was saying?

- myininaya

We can say \[x^3 \approx x^3+1 \]

- JamesJ

The trouble is how exactly is that notation defined? Intuitively, what you've done is right. But can you be sure that it makes sense and doesn't lead to contradictions. The advantage of what Imperialist has done is all the notation he's used is very defined and can be in such a way that we can be sure of our conclusions.

- watchmath

if you are expecting this problem to be answered elementary my suggestion is to split it into computing \(\lim x(\sqrt{x^2+1}-x)\) and \(\lim x(x-\sqrt{x^3+1})\)

- JamesJ

cube root in that second expression, right?

- watchmath

yes, sorry for that

- JamesJ

that's a nice idea. wish I'd thought of it.

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