watchmath
  • watchmath
Compute \[\lim_{x\to +\infty} x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
even with the "x" out front?
watchmath
  • watchmath
yes, the problem is correct
anonymous
  • 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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anonymous
  • anonymous
oh i meant with the x out front it seemed unlikely to be 0
anonymous
  • anonymous
ishaan it is a minus sign. with a plus you would get infinity for sure
anonymous
  • anonymous
it is minus
TuringTest
  • 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
  • TuringTest
oh, ishaan, whoever it was lol
anonymous
  • anonymous
Oh yeah, sorry. thanks satellite
anonymous
  • 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
  • anonymous
that is just a guess though
anonymous
  • anonymous
*Bookmark*
anonymous
  • anonymous
too much algebra for me, i can't do it
Mertsj
  • Mertsj
You can. You just don't want to.
myininaya
  • myininaya
i think i might have it
watchmath
  • watchmath
hello myininaya. Still remember me? :)
myininaya
  • myininaya
yes i do i'm getting the limit is infinity is that right?
myininaya
  • myininaya
so i will show you what i did...
anonymous
  • anonymous
@watchmath myininaya and i were just mentioning you the other day. sunday
anonymous
  • anonymous
i said i missed your problems (not that i can do them) and no the answer is not infinity, but i cheated.
watchmath
  • watchmath
no, the limit is finite
anonymous
  • anonymous
i think the method i wrote would work if i could do the algebra
myininaya
  • 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
  • anonymous
@myininaya your numerator is wrong
anonymous
  • 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
  • myininaya
satellite i disagree i think my numerator is right
anonymous
  • anonymous
oh i am sorry, yes it is right, but you still can't get the limit that way
anonymous
  • 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
  • myininaya
imperialist is a showoff
anonymous
  • anonymous
pretty damned nice
watchmath
  • watchmath
imperialist is correct. Any elementary solution?
anonymous
  • 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
  • anonymous
well i am still thinking of writing the difference of two sixth powers, the algebra is killing me
anonymous
  • 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
  • Mertsj
Why don't you brainiacs go help pre-algebra? He's been trying to get help most of the day.
myininaya
  • myininaya
you know numerically my top approaches 1
myininaya
  • myininaya
algebraically my bottom approaches 2
myininaya
  • 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
  • anonymous
where is the pre-algebra question? i can't find it
Mertsj
  • Mertsj
Updated 41 minutes ago.
myininaya
  • myininaya
So how can I show that \[\lim_{x \rightarrow \infty}(x^2+1-(x^3+1)^\frac{2}{3})=1?\]
anonymous
  • anonymous
Hmm x^2 + 1 - (x^3+1)^(2/3) can't we ignore 1?
myininaya
  • myininaya
couldn't i say \[x^3+1 \approx x^3\]
myininaya
  • myininaya
\[(x^3)^\frac{2}{3}=x^2\]
myininaya
  • myininaya
x^2-x^2=0 we are left with 1
anonymous
  • anonymous
Mhm, sounds good to me!
myininaya
  • myininaya
i think it sounds awesome! :)
watchmath
  • 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
  • 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
  • myininaya
James what do you think about what I was saying?
myininaya
  • myininaya
We can say \[x^3 \approx x^3+1 \]
JamesJ
  • 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
  • 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
  • JamesJ
cube root in that second expression, right?
watchmath
  • watchmath
yes, sorry for that
JamesJ
  • JamesJ
that's a nice idea. wish I'd thought of it.

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