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 ?


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 $x_{n}^2 y_{n} \rightarrow x^2 y$.


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