Question

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

Answer 1

even with the "x" out front?

Answer 2

yes, the problem is correct

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

it is minus

Answer 7

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

Answer 8

oh, ishaan, whoever it was lol

Answer 9

Oh yeah, sorry.
thanks satellite

Answer 10

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
\[(a^6-b^6)=(a-b) (a+b) (a^2-a b+b^2) (a^2+a b+b^2)\]

Answer 11

that is just a guess though

Answer 12

*Bookmark*

Answer 13

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

Answer 14

You can. You just don't want to.

Answer 15

i think i might have it

Answer 16

hello myininaya. Still remember me? :)

Answer 17

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

Answer 18

so i will show you what i did...

Answer 19

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

Answer 20

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

Answer 21

no, the limit is finite

Answer 22

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

Answer 23

\[\lim_{x \rightarrow \infty}\frac{x((x^2+1)^\frac{1}{2}-(x^3+1))^\frac{1}{3}}{1} \cdot \frac{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\]
\[\lim_{x \rightarrow \infty}\frac{x(x^2+1-(x^3+1)^\frac{2}{3})}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\]
\[\lim_{x \rightarrow \infty}\frac{x^3+x-x(x^3+1)^\frac{2}{3}}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\]
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

Answer 24

@myininaya your numerator is wrong

Answer 25

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

Answer 26

satellite i disagree i think my numerator is right

Answer 27

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

Answer 28

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 \[\lim_{x\to \infty}x\left((1+x^2)^\frac12-(1+x^3)^\frac13\right)\] \[=\lim_{x\to\infty}x\left(\left(x+\frac12x^{-1}+O(x^{-3})\right)-\left(x+\frac13x^{-2}+O(x^{-5})\right)\right)\] \[=\lim_{x\to\infty}x\left(\frac12x^{-1}+O(x^{-2})\right)=\lim_{x\to\infty}\frac12+O(x^{-1})=\frac12\]

Answer 29

imperialist is a showoff

Answer 30

pretty damned nice

Answer 31

imperialist is correct. Any elementary solution?

Answer 32

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!

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

you know numerically my top approaches 1

Answer 37

algebraically my bottom approaches 2

Answer 38

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

Answer 39

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

Answer 40

Updated 41 minutes ago.

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

Mhm, sounds good to me!

Answer 47

i think it sounds awesome! :)

Answer 48

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.

Answer 49

(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.)

Answer 50

James what do you think about what I was saying?

Answer 51

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

Answer 52

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.

Answer 53

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 \(\lim x(x-\sqrt{x^3+1})\)

Answer 54

cube root in that second expression, right?

Answer 55

yes, sorry for that

Answer 56

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