First, you can try and prove some important yet simple facts about limits. The following are all equivalent:
$$\eqalign{
& \mathop {\lim }\limits_{x \to a} f\left( x \right) = l \cr
& \mathop {\lim }\limits_{h \to 0} f\left( {a + h} \right) = l \cr
& \mathop {\lim }\limits_{h \to 0} f\left( {a- h} \right) = l \cr
& \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - l} \right] = 0 \cr} $$
You have
$$\mathop {\lim }\limits_{x \to 0} \sqrt {4 - x} + 2 = 4$$
or
$$\mathop {\lim }\limits_{h \to 0} \sqrt {4 - h} + 2 = 4$$
By the above, this is equivalent to
$$\mathop {\lim }\limits_{x \to 4} \sqrt x + 2 = 4$$
So you want to show that for each $\epsilon >0$ there is a $\delta>0$ such that for all $x$, $$0 < \left| {x - 4} \right| < \delta \Rightarrow \left| {\sqrt x - 2} \right| < \varepsilon $$
But multiplying the conjugate gives $$\left| {\sqrt x - 2} \right| = \left| {\frac{{x - 4}}{{\sqrt x + 2}}} \right| < \left| {x - 4} \right|$$
so taking $\delta=\epsilon$ does it.
NOTE You could've also worked with $$\mathop {\lim }\limits_{x \to 0} \sqrt {4 - x} + 2 = 4$$
in fact,
$$\left| {\sqrt {4 - x} + 2 - 4} \right| = \left| {\sqrt {4 - x} - 2} \right| = \left| {\frac{{ - x}}{{\sqrt {4 - x} + 2}}} \right| < \left| x \right|$$
so again, $\delta=\epsilon$, as expected.
The limit of a function $f$ at $-\infty$ is $\ell$ if and only if
$$
\forall\varepsilon\gt0,\ \exists x,\ \forall t,\ t\lt x\implies|f(t)-\ell|\lt\varepsilon.
$$
This can artificially be rewritten in the epsilon-delta frame as
$$
\forall\varepsilon\gt0,\ \exists \delta\gt0,\ \forall t,\ t\lt-1/\delta\implies|f(t)-\ell|\lt\varepsilon.
$$
Best Answer
To show at least the second problem, assume that $\lim_{x\to 0} f(x) = L$. We will prove that $\lim_{x \to a}f(x-a) = L$.
By definition, we are to show : for all $\epsilon > 0$, there exists $\delta > 0$ such that \begin{equation} |y - a |< \delta \implies |f(y-a) - L| < \epsilon \tag{1} \end{equation}
Fix one such $\epsilon$, say $\epsilon_0$.
We know that $\lim_{x \to 0} f(x) = L$. This means, that for the above $\epsilon_0$, there exists some $\delta_0 > 0$ such that \begin{equation}|x - 0|( = |x|) < \delta_0 \implies |f(x) - L| < \epsilon_0 \tag{2}\end{equation}
By taking $\delta = \delta_0$, we see that substituting $y-a = x$ in $(2)$ gives: \begin{equation} |y-a| < \delta_0 \implies |f(y-a) - L |< \epsilon_0 \end{equation}
which is seen to be equivalent to statement $(1)$. Hence, $\lim_{x \to 0} f(x) = \lim_{x \to a} f(x-a)$ is true.
I urge you to prove that if one of these limits does not exist, then the other also doesn't. This involves negation of the definition of limit i.e. what does it mean for $\lim_{x \to a} f(x) \neq L$ for some constant $L$.