Prove that $\lim\limits_{n\to \infty} ⟨k_n, m_n⟩ = ⟨c, d⟩$. Let ${k_n}$ and ${m_n}$ be sequences in $\mathbb{R}^d$ such that $\lim\limits_{n\to \infty} k_n = c$ and $\lim\limits_{n\to \infty} m_n = d$. Prove that $\lim\limits_{n\to \infty} ⟨k_n, m_n⟩ = ⟨c, d⟩$.
My thinking:
$$\lim_{n\to \infty} k_n m_n = cd$$
$$|k_nm_n-cd|<\epsilon$$
$$||k_nm_n||-||cd||\leq|k_nm_n-cd|<\epsilon$$
then ?
edit. $⟨k_n, m_n⟩ = ⟨c, d⟩$ are inner product.
 A: Suppose $\left( \mathbf{a_n} \right)_{n \in \mathbb{N}}$ and $\left( \mathbf{b_n} \right)_{n \in \mathbb{N}}$ are sequences in $\mathbb{R}^k$. 
Then, for each $n \in \mathbb{N}$, we have
$$ \mathbf{a}_n, \mathbf{b}_n \in \mathbf{R}^k, $$
and thus
$$ \mathbf{a_n} = \left( \alpha_{n1}, \ldots, \alpha_{nk} \right) $$
and
$$ \mathbf{b_n} = \left( \beta_{n1}, \ldots, \beta_{nk} \right), $$
where
$$ \alpha_{ni}, \beta_{ni} \in \mathbb{R} $$
for  $i = 1, \ldots, k$.
Then, for each $n \in \mathbb{N}$, we have
$$ \left\langle \mathbf{a}_n, \mathbf{b}_n \right\rangle = \sum_{i=1}^k \alpha_{nk}\beta_{nk}. $$
Now let
$$  \mathbf{a} = \left( \alpha_1, \ldots, \alpha_k \right) $$
and
$$  \mathbf{b} = \left( \beta_1, \ldots, \beta_k \right) $$
be any elements of $\mathbb{R}^k$.
Then 
$$ \lim_{n \to \infty} \mathbf{a}_n = \mathbf{a} $$
if and only if
$$ \lim_{n \to \infty} \alpha_{ni} = \alpha_i $$
for each $i = 1, \ldots, k$.
And, similarly, 
$$ \lim_{n \to \infty} \mathbf{b}_n = \mathbf{b} $$
if and only if
$$ \lim_{n \to \infty} \beta_{ni} = \beta_i $$
for each $i = 1, \ldots, k$.
Therefore if 
$$ \lim_{n \to \infty} \mathbf{a}_n = \mathbf{a} \ \mbox{ and } \ \lim_{n \to \infty} \mathbf{b}_n = \mathbf{b}, $$
then we have
$$ \lim_{n \to \infty} \alpha_{ni} = \alpha_i \ \mbox{ and } \  \lim_{n \to \infty} \beta_{ni} = \beta_i $$
for each $i = 1, \ldots, k$.
Hence we have 
$$
\begin{align}
 \lim_{n \to \infty} \left\langle \mathbf{a}_n, \mathbf{b}_n \right\rangle &= \lim_{n \to \infty} \left( \sum_{i=1}^k \alpha_{ni}\beta_{ni} \right) \\
&=  \sum_{i=1}^k \left[ \lim_{n \to \infty} \left( \alpha_{ni} \beta_{ni} \right) \right] \\
&= \sum_{i=1}^k \left[  \left( \lim_{n \to \infty}  \alpha_{ni} \right) \left( \lim_{n \to \infty} \beta_{ni} \right) \right] \\
&= \sum_{i=1}^k \alpha_i \beta_i \\
&= \langle \mathbf{a}, \mathbf{b} \rangle.
\end{align}
$$
Hope this helps.
Should have any confusion, or further questions, please feel free to inquire.
