1. ## More Functional Limits

(1) Let $g: A \rightarrow \mathbb{R}$ and assume that $f$ is a bounded function on $A \subseteq \mathbb{R}$ (i.e., there exists $M > 0$ satisfying $|f(x)| \leq M$ for all $x \in A$). Show that if $lim_{x \rightarrow c}$ $g(x)=0$, then $lim_{x \rightarrow c}$ $g(x)f(x)=0$ as well.

(2) Assume $f(x) \geq g(x)$ for all $x$ in some set $A$ on which $f$ and $g$ are defined. Show that for any limit point $c$ of $A$ we must have $lim_{x \rightarrow c}$ $f(x)$ $\geq$ $lim_{x \rightarrow c}$ $g(x)$.

Thanks for looking and any help I can be provided! I am really unsure of how to approach these at all.

(1) Let $g: A \rightarrow \mathbb{R}$ and assume that $f$ is a bounded function on $A \subseteq \mathbb{R}$ (i.e., there exists $M > 0$ satisfying $|f(x)| \leq M$ for all $x \in A$). Show that if $lim_{x \rightarrow c}$ $g(x)=0$, then $lim_{x \rightarrow c}$ $g(x)f(x)=0$ as well.
Give condition implies:

$\forall \epsilon > 0 ,\exists \delta > 0: |x - c| < \delta \Rightarrow |g(x) - 0| < \epsilon$

That means if we choose $\epsilon$ as $\frac{\epsilon}{M}$, there exists a delta such that $|x - c| < \delta \Rightarrow |g(x) - 0| < \frac{\epsilon}{M}$.

Choose this delta to prove continuity of f(x)g(x)...

More precisely, we have to prove that:

$\forall \epsilon > 0 ,\exists \delta > 0: |x - c| < \delta \Rightarrow |f(x)g(x) - 0| < \epsilon$

For any $\epsilon$ choose the $\delta$said previously,

Now $|f(x)g(x) - 0| = |f(x)g(x)|

3. Perfect! Thanks!

(2) Assume $f(x) \geq g(x)$ for all $x$ in some set $A$ on which $f$ and $g$ are defined. Show that for any limit point $c$ of $A$ we must have $lim_{x \rightarrow c}$ $f(x)$ $\geq$ $lim_{x \rightarrow c}$ $g(x)$.
Assume the contrary, that's $\lim_{x\rightarrow{c}}f(x)<\lim_{x\rightarrow{c}}g (x)$

Or equivalently: $0<\lim_{x\rightarrow{c}}[g(x)-f(x)]=d$

By definition: $
\forall \varepsilon > 0,\exists \delta _\varepsilon > 0/{\text{ if }}0 < \left| {x - c} \right| < \delta _\varepsilon {\text{ we have:}}{\text{ }}\left| {g\left( x \right) - f\left( x \right) - d} \right| < \varepsilon
$

Choose: $
\varepsilon = \tfrac{d}
{2}
$
thus we have: $
{\text{if }}0 < \left| {x - c} \right| < \delta _{\tfrac{d}
{2}} \Rightarrow 0 < \tfrac{1}
{2} \cdot d < g\left( x \right) - f\left( x \right) < \tfrac{3}
{2} \cdot d
$

And this is absurd since it goes against the hypothesis $
f\left( x \right) > g\left( x \right)
$
for all $
x \in A
$