Whether the sequence $\{x_n-y_n\}$ converge or not. Given, $\{x_n\}$ and $\{y_n\}$ are two real valued sequences such that $\lim_{n\to\infty} (x_{n}^{3}-y_{n}^{3})=0 $
Then is it always true that $\lim_{n\to\infty} (x_n-y_n) =0 $ ?
I tried by splitting $ (x_{n}^{3}-y_{n}^{3}) $, but that didn't help as $\lim_{n\to \infty} (x_{n}^{2}+x_{n}y_{n}+y_{n}^{2}) $ can be $0$ or non zero or doesn't exist.
 A: Suppose that $\lim_n(x_n-y_n)\ne 0$; then there are subsequences $\langle x_{n_k}:k\in\Bbb N\rangle$ and $\langle y_{n_k}:k\in\Bbb N\rangle$ and an $\epsilon>0$ such that $|x_{n_k}-y_{n_k}|\ge\epsilon$ for all $k\in\Bbb N$. Infinitely many of the differences $x_{n_k}-y_{n_k}$ must have the same sign, so without loss of generality we may assume that $x_{n_k}-y_{n_k}\ge\epsilon$ for all $k\in\Bbb N$. Then $x_{n_k}\ge y_{n_k}+\epsilon$ for all $k\in\Bbb N$, and hence
$$\begin{align*}
x_{n_k}^3-y_{n_k}^3&\ge(y_{n_k}+\epsilon)^3-y_{n_k}^3\\
&=\epsilon\left(3y_{n_k}^2+3\epsilon y_{n_k}+\epsilon^2\right)\\
&=\epsilon\left(3\left(y_{n_k}+\frac{\epsilon}2\right)^2+\frac{\epsilon^2}4\right)\\
&\ge\frac{\epsilon^3}4
\end{align*}$$
for all $k\in\Bbb N$, contradicting the hypothesis that $\lim_n(x_n^3-y_n^3)=0$.
A: Yes, $x_n^3 -y_n^3 \to 0 $ implies $ x_n - y_n \to 0$ for real-valued sequences. This follows from the following estimate:

For all $x, y \in \Bbb R$ is $|x-y| \le \sqrt[3]{4|x^3 - y^3|}$.

(In other words: the function $x \mapsto x^{1/3}$ is “Hölder continuous” with exponent $1/3$.)
Due to the symmetry, it suffices to prove the inequality for the case $x \ge y$:
$$
\begin{align}
 4(x^3-y^3) &= 4(x-y)(x^2+xy+y^2) \\
&= (x-y)\left( (x-y)^2 + 3(x+y)^2\right) \\
&\ge (x-y)^3 \, .
\end{align}
$$
An alternative proof (which generalizes to other odd exponents) would be to compute the maximum of the function
$$
 f(u) = \frac{(u-1)^3}{u^3-1} = \frac{(u-1)^2}{u^2+u+1}
$$
which turns out to be $f(-1) = 4$.
