Question

Compute the limit or show that the limit does not exist

Answer 1

\[\lim_{x \rightarrow \infty}\sin (\sqrt[3]{2012+x})-\sin(\sqrt[3]{x})\]

Answer 2

is it the 3rd root or the 9th root?

Answer 3

the 3rd root

Answer 4

Would it make sense to show that:\[\lim_{n\rightarrow \infty} \sqrt[3]{x+2012}-\sqrt[3]{x}=0\]So that for sufficiently large x, the two are basically the same?

Answer 5

Then we can claim that the limit in question is 0?

Answer 6

\[-1\le sinx \le1\] i think not the same

Answer 7

I feel quite certain that the limit exists and is 0. I think if you can show that cuberoot(x+2012) and cuberoot(x) get arbitrarily close, then you're basically there.

Answer 8

I'm having trouble formally proving it though...
fix epsilon >0
show that there exists A, such that for all x>= A,
\[|\sqrt[3]{x+2012} - \sqrt[3]{x}| < \epsilon\]

Answer 9

@SmoothMath, don't ignore that the second term is a Sin function and the value can only be -1 thru 1

Answer 10

Both terms are sin functions.

Answer 11

multiply numerator and denominator by
a^2 + ab + b^2 --> a^3 - b^3

Answer 12

Yes, they both sin functions oscillate between 1 and -1, and they both have a really ugly non-constant period. However, as joemath showed, the arguments for the sin functions get arbitrarily close to each other. So the results of the sin functions get arbitrarily close, their difference approaches 0.

Answer 13

Was it necessary that I expand:\[\sin (x^{\frac{1}{3}}+\epsilon )\] or could I have just let epsilon go to 0 from there?

Answer 14

Stating it formally is kind of a pain though...
Joe's seems pretty close, but it's not the way I've been taught to give this kind of proof =/
Seems like we need to fix some arbitrarily small epsilon and then give a sufficient value A for which all x>A gives f(x) <epsilon-0, which is just f(x) < epsilon...

Answer 15

I was ignoring the fact the first term is also a sin function sorry about that.

Answer 16

I did the same at first.

Answer 17

When i first looked at it, i didnt see any sin functions at all lol >.<

Answer 18

Since the lim without the sin functions goes to 0, dont we know there exists a real number A such that x>A implies the difference is less than epsilon?

Answer 19

Intuitively, yes... Can't seem to get that result though.

Answer 20

http://en.wikipedia.org/wiki/Cauchy_sequence
something like this??

Answer 21

I guess technically the problem statement only asks you to compute the limit or prove that it doesn't exist. Since we compute it to be 0, we don't necessarily have to prove that. Hardly satisfying though.

Answer 22

well .. perhaps .. we could use mean value theorem,

\[ |f(x) - f(a)| <= (x-a) \times some\; finite \; vlaue\]

Answer 23

Is it true?

Answer 24

i guess it is ...

Answer 25

I am not sure if you can call it proof
Using Mean value theorem, ... \( y \) belongs to (x, x+2012)
\[ \sin (\sqrt[3]{2012+x} - \sin (\sqrt[3]{x}| = 2012 \frac{\cos\sqrt[3]{y}}{3(y)^{2/3} }\leq 2012 \left | \frac{\cos\sqrt[3]{y}}{3(y)^{2/3} } \right |\]
\[ \lim_{x \rightarrow \infty}|\sin (\sqrt[3]{2012+x} - \sin (\sqrt[3]{x}| < \epsilon \]