Prove using the definition of a limit, that $f(x) >$ something if $|x| < \delta$ The function $f (x)$ is defined for $−∞ < x < ∞$. In addition, we have
$$\lim_{x \to 0} f(x) = 2$$
(a) Give the $\epsilon$-$δ$-definition of $\lim_{x \to 0} f(x) = 2$.
(b) Prove (using this definition) that a number $δ > 0$ exists such that
$f(x) > \frac{3}{4}$ if $|x| < δ$.

So the first part is easy:
$ \forall\epsilon>0\ \exists\delta>0$ such that if $0<|x-0|<\delta$ then $|f(x)-2|<\epsilon$
The second part is where I'm having trouble. I know that when $f(x)>\frac{3}{4}$ we are at a point where $\frac{5}{4}<\epsilon$.
How do I proceed?
 A: Hint: consider $\epsilon=1/4$; then there exists $\delta>0$ such that
$$
|f(x)-2|<\frac{1}{4}
$$
for $0<|x|<\delta$. This means
$$
-\frac{1}{4}<f(x)-2<\frac{1}{4}
$$
hence
$$
f(x)>2-\frac{1}{4}
$$
for $0<|x|<\delta$. Note the $0<|x|$ clause.
This is as much as you can prove; the statement, as is, is incorrect because the function
$$
f(x)=\begin{cases}
2&\text{for $x\ne0$}\\
-42&\text{for $x=0$}
\end{cases}
$$
satisfies the hypotheses, but there's no $\delta>0$ such that $f(x)>\frac{3}{4}$ for $|x|<\delta$.
A: As you have shown in part (a), there exists some $\delta > 0$ such that $|f(x)-2|$ can be as small as we want, that is, $f(x)$ can be as close to $2$ as we want. How close do we need so that $f(x) > 3/4$? Note that if $f(x)$ is a distance of one or less from $2$, then $f(x)$ will be greater than or equal to $1$, and so also greater than $3/4$. So we choose $\epsilon = 1$, and part (a) gives:
$$
|x|<\delta \quad \Longrightarrow \quad |f(x)-2|<1.
$$
Then
$$
|f(x)-2|<1 \quad \Longrightarrow \quad -1 < f(x) - 2 < 1 \quad \Longrightarrow \quad 1 < f(x) < 3 \quad \Longrightarrow \quad f(x) > 1 > \frac{3}{4}.
$$
