Proof through definition of convergence 
Let $(x_n)_{n=1}^\infty$ and $(y_n)_{n=1}^\infty$ be two sequences of real numbers and $l\in \mathbb{R}$. Assume that $\lim_{n \to \infty} x_{n}=0$ and that there exists $n_{0} \in \mathbb{N}$ such that for all $n \geq n_{0}$ we have $|y_{n}-l| \leq |x_{n}|$.  Show that $\lim_{n \to \infty} y_{n}=l$.

What I have is  $\lim_{n \to \infty} |y_{n}| \leq \lim_{n \to \infty} |x_{n}|$ and  $\lim_{n \to \infty} y_{n} - \lim_{n \to \infty} l \leq 0$ so  $\lim_{n \to \infty} y_{n} -l \leq 0$ or  $\lim_{n \to \infty} y_{n} \leq l$. But by the definition of convergence,   $\lim_{n \to \infty} y_{n}=l$. Can I use the convergence definition this way? 
 A: From what you wrote it is hard for me to decode your intentional meaning. 
It is actually very simple: If $\varepsilon > 0$, then by assumption there is some $N$ such that $|x_{n}| < \varepsilon$ for all $n \geq N$. By assumption there is some $n_{0}$ such that $|y_{n}-l| \leq |x_{n}|$ for all $n \geq n_{0}$; so $n \geq \max \{ n_{0}, N \}$ implies $|y_{n}-l| \leq |x_{n}| < \varepsilon$. This shows that $y_{n} \to l$.
A: I would suggest the hint of using the triangle inequality:
$$
\begin{align}
|y_{n} - l| < |x_{n}| &\Rightarrow -x_{n} < y_{n} - l < x_{n}\\
|x_{n}| < \epsilon &\Rightarrow -\epsilon < x_{n} < \epsilon
\end{align}
$$
where the second inequality can be made true for sufficiently large $n$ for arbitrary $\epsilon$ by the definition of convergence for $x_{n}$. Of course the first is true for sufficiently large $n > N_{0}$ by the information in the question.
Can you see how to combine these facts together to show the $y_{n}$ has limit $l$ by showing this hypothesis satisfies the definition of convergence? 
