Finding the limit is not equal to something Prove that $\lim_{x\to 3}(4-3x)\neq -4.9$.
I know that I need to prove the negation of the definition:
$\exists \epsilon >0, \forall \delta > 0, \exists x \in \Bbb {R}, ((|x-a| < \delta \wedge x \neq a) \wedge |f(x) -L| \geq \epsilon)$
So I fixed delta but I don't know how to find the value for epsilon.  And then I also need to restrict a $\delta_2$ here?
 A: We'll choose $\epsilon=\left|\dfrac{-5-(-4.9)}2\right|=0.05$ since there is a $\delta_0$ such that for all $x$ within $\delta_0$ of $3$ (and not at $3$), $f(x)$ is within $0.05$ of $-5$ (this follows from $\lim_{x\to 3}(4-3x)=-5$) and thus is not within $0.05$ of $-4.9$.
Note the above statement is sufficient to prove your negated statement (can you see why?). If you try to prove the limit, you'll find $\delta_0=\dfrac{\epsilon}3=\dfrac{0.05}3$.
Thus, we know for all $x$ such that $0<|x-3|<\delta_0=\dfrac{0.05}{3}$, $|f(x)-4.9|\geq 0.05$.
Let $\delta>0$.
Clearly for any $\delta\geq\delta_0$, we can choose any one of the $x$'s that have the property $0<|x-3|<\delta_0=\dfrac{0.05}{3}$. That is $3-\dfrac{0.05}3<x<3+\dfrac{0.05}3$ and $x\neq3$.
If $\delta<\delta_0$, then since we are already within $\delta_0$, we can choose any $x$ such that $0<|x-3|<\delta$.

Proof:
Choose $\epsilon=0.05$. Let $\delta>0$.
Case $1$: $\delta\geq\dfrac{0.05}3$. Choose $x=3.01$. 
Then $|x-3|=|3.01-3|=0.01<\dfrac{0.05}{3}\leq \delta$ and $x=3.01\ne3$.
We have $|f(x)-(-4.9)|=|8.9-3x|=|8.9-3(3.01)|=0.13\geq0.05=\epsilon$.
Case $2$: $\delta<\dfrac{0.05}3$. Choose $x=3+\dfrac{\delta}2$.
Then $|x-3|=\left|\dfrac{\delta}2\right|<\delta$ and $x=3+\dfrac{\delta}2\ne3$.
We have \begin{align}&|f(x)-(-4.9)|=|8.9-3x|=\left|\dfrac{3\delta}{2}-0.1\right|
\\=&0.1-\dfrac{3\delta}{2} \hspace{50ex}(\;\because \frac32\delta<\frac32\frac{0.05}3<0.1)
\\>&0.1-\dfrac{3}2\dfrac{0.05}3=0.075\geq0.05=\epsilon.\end{align}

Alternatively, you may prove $\lim_{x\to3} 4-3x=-5$ and thus, conclude your statement.
A: 
I going to use the hint given above. Let $\varepsilon_0=0.05$ and $\delta>0$ arbitrary but fix. We have to find such $x$ in the above negation. Let $x=x_0=3+\frac{\delta}{m}$ for any $m>1$ . Then $x_0$ satisfy  $x_0\neq 3$ and $|x_0-3|<\delta$ but $$|f(x_0)+4.9|=\cdots=|-0.1-3\delta/m|>0.1=2\varepsilon_0>\varepsilon_0$$

