# $\lim_{n \rightarrow \infty} (x_{n}^2 y_{n}) = (\lim_{n \rightarrow \infty } x_{n})^2 (\lim_{n \rightarrow \infty } y_{n})$

If $$x_{n},y_{n}$$ are two convergent sequences converging to $$x,y$$ respectively. Then we have to prove $$x_{n}^2 y_{n}$$ converges to $$x^2 y$$ or $$\lim_{n \rightarrow \infty} x_{n}^2 y_{n} = (\lim_{n \rightarrow \infty} x_{n})^2 (\lim_{n \rightarrow \infty} y_{n})$$.

I thought of using triangle inequality $$|x_{n}^2 y_{n} - x^2 y| \leq |y_{n} - y||x_{n}^2| + |x_{n}^2 -x^2||y|$$. As $$x_{n}$$ is convergent that implies $$x_{n}$$ is bounded implying $$|x_{n}| \leq B$$. We want to show $$|x_{n}^2 y_{n} - x^2 y| < \epsilon$$ in order to show the convergence. I am still thinking how to prove it ?

EDIT -

We have $$\lim_{n \rightarrow \infty } x_{n} = x$$ and $$\lim_{n \rightarrow \infty} y_{n} = y$$.

We need to prove that $$\lim_{n \rightarrow \infty} x_{n}^2 y_{n} = x^2y$$.

We use the triangle inequality trick!

Essentially we want to show that there exists $$k \in \Bbb{N}$$ such that $$\forall n \geq k$$, we have $$|x_{n}^2 y_{n} - x^2 y| < \epsilon$$.

$$|x_{n}^2 y_{n} - x^2 y| = |x_{n}^2 y_{n} -x_{n}^2 y + x_{n}^2 y - x^2 y| = |x_{n}^2(y_{n} - y) + y(x_{n}^2 -x^2)| \leq |x_{n}^2| |y_{n} -y| + |y| |x_{n}^2 -x^2|$$

Since $$x_{n}$$ is bounded there exists $$B$$ such that $$|x_{n}| \leq B$$. Let $$M =$$ max $$\{B,||y\}$$.

$$\leq M(|y_{n} -y| + |x_{n}^2| - x^2)$$

Let $$\epsilon > 0$$, choose $$k \in \Bbb{N}$$ such that $$\forall n \geq k$$; we have $$|y_{n} -y| < \frac{\epsilon}{2M}$$ and $$|x_{n}^2 -x^2| < \frac{\epsilon}{2M}$$ implying $$x_{n}^2 y_{n} \rightarrow x^2 y$$.