Proof of limits as one function goes to zero and one is bounded Suppose that $$ \lim_{x\to +\infty} f(x) = 0$$ and $$g(x)$$ is a bounded .
Show that $$\lim_{x\to +\infty}f(x)g(x)=0$$
Thanks in advance 
 A: This can be done by noting that $|g(x)| < M$ so  $|f(x)g(x)| < f(x)\cdot M$. Since $f(x)\to 0$ you conclude $|f(x)g(x)|\to 0$. Then the trick is to argue from here as to why this means $f(x)g(x)\to 0$.
A: $$\lim _{ x\to +\infty  } f(x)=0\Rightarrow \quad \forall x\in R,\exists M>0\quad \left| f\left( x \right)  \right| \le M\\ \forall x\in R,\exists \frac { \epsilon  }{ M } >0\quad \left| g\left( x \right)  \right| \le \frac { \epsilon  }{ M } \\ \left| f\left( x \right) g\left( x \right)  \right| \le \left| f\left( x \right)  \right| \left| g\left( x \right)  \right| \le M\cdot \frac { \epsilon  }{ M } =\epsilon $$ which means $$\lim_{x\to +\infty}f(x)g(x)=0$$
A: Let $\alpha$ be an upper bound of b such that -$\alpha$ is a lower bound for g. Then alpha is a constant. This means that ($\alpha f_x$) converges to 0. 
For each $\epsilon>0$, choose the m that you'd choose for ($\alpha f_x$). Let x be in $\mathbb R$ and m < x. Then $\left |f_x g_x-0 \right |< \left |\alpha f_x-0 \right|<\epsilon$. Thus ($f_xg_x$) converges to 0 as well. 
Q.E.D
Adam V. Nease
