Prove: If $\lim_{x \to \infty}g(x)=0$ $ \Rightarrow \lim_{x \to \infty} f(x)=0$ 
$\forall x \in \mathbb R$ $g(x)\gt 0 $ and $\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)}=L \gt0$
Prove that
1.) If $\displaystyle \lim_{x \to \infty}g(x)=0$ then $\displaystyle \lim_{x \to \infty} f(x)=0$
2.) Conclude that $\displaystyle \lim_{x \to \infty}g(x)=0 \iff \lim_{x \to \infty} f(x)=0$

1.) Using the definition of limits:
$\forall \epsilon_1 \gt 0$, $N_1 \gt 0$
$$ x\gt N_1  \Rightarrow |g(x)| \lt \epsilon_1$$
$\forall \epsilon_0 \gt 0$, $N_0 \gt 0$
$$ x \gt N_0  \Rightarrow \left|\frac{f(x)}{g(x)}-L\right| \lt \epsilon_0$$
Now doing a little bit of algebra:
$$g(x)(-\epsilon_0+L)\lt f(x) \lt g(x)(\epsilon_0+L)$$
And using the fact that $|g(x)| \lt \epsilon_1$:
$$-\epsilon_1(-\epsilon_0+L) \le g(x)(-\epsilon_0+L)\lt f(x) \lt g(x)(\epsilon_0+L)\le \epsilon_1(\epsilon_0+L)$$
This turns into:
$$-\epsilon_1(-\epsilon_0+L) \lt f(x) \lt \epsilon_1(\epsilon_0+L)$$
Are my steps correct thus far? Leaving the only thing left is picking a value for $\epsilon_1$ so that I end up with $|f(x)|\lt \epsilon$?
 A: This is correct. You may choose to fix a value for $\varepsilon_0$ (for example $\varepsilon_0=\dfrac{L}{2}$) and then choose the good value of $\varepsilon_1$ so that $\varepsilon_1(\varepsilon_0+L) \leq \varepsilon$ and $-\varepsilon_1(-\varepsilon_0+L) \geq \varepsilon$.
A: Actually you can use the following way to prove.
1.) Since $\lim_{x\to\infty}g(x)=0$ and $\lim_{x\to\infty}\frac{f(x)}{g(x)}=L$, for $\forall \epsilon>0$, there is $N>0$ such that, for $|x|>N$, one has
$$ |g(x)|<\epsilon/(2L), \big|\frac{f(x)}{g(x)}-L\big|<\epsilon/2. $$
One also can assume that for $|x|>N$, $|g(x)|<1$.
Then for $|x|>N$,
\begin{eqnarray}
|f(x)|&=&\big|(\frac{f(x)}{g(x)}-L)g(x)+Lg(x)\big|\\
&\le& \big|(\frac{f(x)}{g(x)}-L)g(x)\big|+|L||g(x)|\\
&<& \epsilon/2+\epsilon/2\\
&=&\epsilon.
\end{eqnarray}
So 
$$\lim_{x\to\infty}f(x)=0. $$
2.) Assume $\lim_{x\to\infty}f(x)=0$. If $\lim_{x\to\infty}g(x)\not=0$, then
$$ \lim_{x\to\infty}\frac{f(x)}{g(x)}=0 $$
which is against
$$ \lim_{x\to\infty}\frac{f(x)}{g(x)}=L\not= 0.$$
A: It is much simpler / better to use the laws of algebra of limits which are much more powerful tools than people actually think. An appeal to the definition of limit involving $\epsilon, \delta$ should be avoided unless it is necessary (or explicitly asked by examiner).
Thus for the first problem we see that if $\lim_{x \to \infty}g(x) = 0$ then $$\lim_{x \to \infty}f(x) = \lim_{x \to \infty}\frac{f(x)}{g(x)}\cdot g(x) = L \cdot 0 = 0$$
Next problem asks us to prove the converse and here we must have $L \neq 0$ otherwise this converse does not hold. Well, we are given in the question that $L > 0$. Let $\lim_{x \to \infty}f(x) = 0$ and then $$\lim_{x \to \infty}g(x) = \lim_{x \to \infty}\frac{f(x)}{f(x)/g(x)} = \frac{0}{L} = 0$$ and combining this result with the one in previous paragraph we have the following result:
If $\lim\limits_{x \to \infty}\dfrac{f(x)}{g(x)} = L \neq 0$ then $$\lim_{x \to \infty}g(x) = 0\Longleftrightarrow \lim_{x \to \infty}f(x) = 0$$
From the above you can also observe that the conditions $g(x) > 0$ and $L > 0$ are unnecessary. Only requirement is that $\lim_{x \to \infty}f(x)/g(x)$ exists and is non-zero and your conclusions in the question remain valid.
