I must demonstrate that if $\displaystyle \{ x_n \}$, $\displaystyle \{ y_n \}$ and $\displaystyle \{ \alpha _n \}$ are sequences of real number then the following is true:

if $\displaystyle x_n=o( \alpha _n)$ then $\displaystyle x_n=O(\alpha _n )$.

My attempt: I must show that knowing that $\displaystyle \lim _{n \to \infty} \frac{x_n}{\alpha _n}=0$ then there exist a constant $\displaystyle C$ and an integer $\displaystyle r$ such that $\displaystyle |x_n|<C|\alpha _n|$ for $\displaystyle n>r$.

But to me, any constant in R that I'd take would make $\displaystyle \lim _{n \to \infty} \frac{x_n}{\alpha _n} \neq 0$. For instance if the sequence $\displaystyle \alpha_n =x_n$, then $\displaystyle \lim _{n\to \infty} \frac{x_n}{\alpha _n}=\frac{1}{C} \neq 0$. In fact C would have to be infinite for the relation to be true.

Am I missing something or the problem asks me to show a wrong result?