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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0even with the "x" out front?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0yes, the problem is correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\to +\infty} x\times \frac{x}{x}\left(\sqrt{x^2+1}\sqrt[3]{x^3+1}\right)\] \[ x^2 ( \sqrt{1 + \frac{1}{x^2}} + \sqrt[3]{1 + \frac{1}{x^3}})\] I think it must be \(\infty\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i meant with the x out front it seemed unlikely to be 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ishaan it is a minus sign. with a plus you would get infinity for sure

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0lost 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh, ishaan, whoever it was lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, sorry. thanks satellite

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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 \[(a^6b^6)=(ab) (a+b) (a^2a b+b^2) (a^2+a b+b^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is just a guess though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0too much algebra for me, i can't do it

Mertsj
 5 years ago
Best ResponseYou've already chosen the best response.0You can. You just don't want to.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i think i might have it

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0hello myininaya. Still remember me? :)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0yes i do i'm getting the limit is infinity is that right?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so i will show you what i did...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@watchmath myininaya and i were just mentioning you the other day. sunday

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i said i missed your problems (not that i can do them) and no the answer is not infinity, but i cheated.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0no, the limit is finite

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think the method i wrote would work if i could do the algebra

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty}\frac{x((x^2+1)^\frac{1}{2}(x^3+1))^\frac{1}{3}}{1} \cdot \frac{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\] \[\lim_{x \rightarrow \infty}\frac{x(x^2+1(x^3+1)^\frac{2}{3})}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\] \[\lim_{x \rightarrow \infty}\frac{x^3+xx(x^3+1)^\frac{2}{3}}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0@myininaya your numerator is wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0middle 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 \[(a^6b^6)=(ab) (a+b) (a^2a b+b^2) (a^2+a b+b^2)\]to clear them

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0satellite i disagree i think my numerator is right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i am sorry, yes it is right, but you still can't get the limit that way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here'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
 5 years ago
Best ResponseYou've already chosen the best response.0imperialist is a showoff

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0imperialist is correct. Any elementary solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0well i am still thinking of writing the difference of two sixth powers, the algebra is killing me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Why don't you brainiacs go help prealgebra? He's been trying to get help most of the day.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you know numerically my top approaches 1

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0algebraically my bottom approaches 2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0if some how i can show without a numerical approach or graphical approach that the top approaches 1, then I would be dandy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where is the prealgebra question? i can't find it

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0So how can I show that \[\lim_{x \rightarrow \infty}(x^2+1(x^3+1)^\frac{2}{3})=1?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm x^2 + 1  (x^3+1)^(2/3) can't we ignore 1?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0couldn't i say \[x^3+1 \approx x^3\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0\[(x^3)^\frac{2}{3}=x^2\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0x^2x^2=0 we are left with 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mhm, sounds good to me!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i think it sounds awesome! :)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0(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
 5 years ago
Best ResponseYou've already chosen the best response.0James what do you think about what I was saying?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0We can say \[x^3 \approx x^3+1 \]

JamesJ
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0cube root in that second expression, right?

JamesJ
 5 years ago
Best ResponseYou've already chosen the best response.0that's a nice idea. wish I'd thought of it.
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