If $\lim_{i \to \infty} \frac{x_i}{y_i} = 0$ and $\lim_{i \to \infty} y_i = K$ with $K>0$, $\lim_{i \to \infty} x_i = 0$? If $$\lim_{i \to \infty} \frac{x_i}{y_i} = 0$$ and $$\lim_{i \to \infty} y_i = K$$ with $K>0$ and $K$ real, $x_i$, $y_i$ real, will it be the case that $$\lim_{i \to \infty} x_i = 0$$?
 A: Yes. Indeed for large $i$ $$|x_i| = \Big|\frac{x_i}{y_i}\Big||y_i| \le \Big|\frac{x_i}{y_i}\Big|\cdot 2K \to 0.$$
A: You can use your product limit laws here, you know that when $\lim a_n, \lim b_n$ both exist that
$$\lim a_nb_n = (\lim a_n)(\lim b_n)$$
here let $a_n = {x_n\over y_n}, b_n = y_n$.
Then
$$\lim x_n = \lim {x_n\over y_n}\cdot y_n = 0 \cdot K=0.$$
A: Let's go back to the limit definitions:
$\lim_{i \to \infty} \frac{x_i}{y_i} = 0 \Leftrightarrow \forall \varepsilon >0, \exists N \in \mathbb N, i>N \implies |\frac {x_i} {y_i}| < \varepsilon $
$\lim_{i\to \infty} y_i = K \Leftrightarrow \forall \varepsilon > 0, \exists M \in \mathbb N, i>M \implies |y_i -K| < \varepsilon$
Let's take $\varepsilon > 0$. As $K>0$, we have $\varepsilon ' = min(\frac \varepsilon {2K}, K)>0$. From the above definitions, we can find a $N$ and $M$ verifying the properties for $\varepsilon '$. Let us take $L=max(N, M)$
$i>L \implies K-\varepsilon ' < y_i < K+ \varepsilon ' \implies 0<K-\varepsilon ' < y_i < K+ \varepsilon '$
$\begin{align}i > L \implies |\frac{x_i}{y_i}| < \varepsilon ' &\implies |x_i|<\varepsilon ' y_i \le \varepsilon ' (K + \varepsilon') \le \frac \varepsilon {2K} ( K+ K) \\
&\implies |x_i| < \varepsilon\end{align}$
So we have shown $\forall \varepsilon > 0, \exists L \in \mathbb N, i>L \implies |x_i| < \varepsilon$, which proves $\lim_{i\to \infty} x_i = 0$
