# Thread: More Functional Limits

1. ## More Functional Limits

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

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

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

2. Originally Posted by shadow_2145
(1) Let $\displaystyle g: A \rightarrow \mathbb{R}$ and assume that $\displaystyle f$ is a bounded function on $\displaystyle A \subseteq \mathbb{R}$ (i.e., there exists $\displaystyle M > 0$ satisfying $\displaystyle |f(x)| \leq M$ for all $\displaystyle x \in A$). Show that if $\displaystyle lim_{x \rightarrow c}$ $\displaystyle g(x)=0$, then $\displaystyle lim_{x \rightarrow c}$ $\displaystyle g(x)f(x)=0$ as well.
Give condition implies:

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

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

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

More precisely, we have to prove that:

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

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

Now $\displaystyle |f(x)g(x) - 0| = |f(x)g(x)| <M\frac{\epsilon}{M}= \epsilon$

3. Perfect! Thanks!

4. Originally Posted by shadow_2145
(2) Assume $\displaystyle f(x) \geq g(x)$ for all $\displaystyle x$ in some set $\displaystyle A$ on which $\displaystyle f$ and $\displaystyle g$ are defined. Show that for any limit point $\displaystyle c$ of $\displaystyle A$ we must have $\displaystyle lim_{x \rightarrow c}$ $\displaystyle f(x)$ $\displaystyle \geq$ $\displaystyle lim_{x \rightarrow c}$ $\displaystyle g(x)$.
Assume the contrary, that's $\displaystyle \lim_{x\rightarrow{c}}f(x)<\lim_{x\rightarrow{c}}g (x)$

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

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

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

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