1. ## Limits

1. $
Suppose \mid f(x) \mid \leq \mid \frac{B}{x} \mid for\ all\ x\ not\ equal\ to\ 0.$

$Show\ that:
\lim_{x \to 0} ~ x^2f(x)=0
$

2. $
Suppose\ that\ \lim_{x \to 0} ~ f(x)=0\ and\ that\ \forall x \in R\ g(x) \leq B.$

$Show \lim_{x \to 0} ~ f(x)g(x)=0
$

Sorry, I don't even know where to start..

2. Originally Posted by Pisces
1. $
Suppose \mid f(x) \mid \leq \mid \frac{B}{x} \mid for\ all\ x\ not\ equal\ to\ 0.$

$Show\ that:
\lim_{x \to 0} ~ x^2f(x)=0
$
Note, $|x^2f(x)| \leq x^2\left| \frac{B}{x} \right| = B|x|$.
But $\lim_{x\to 0}B|x| = 0$.
Thus, by sequeeze theorem we have $\lim_{x\to 0}x^2f(x) = 0$.

2. $
Suppose\ that\ \lim_{x \to 0} ~ f(x)=0\ and\ that\ \forall x \in R\ g(x) \leq B.$

$Show \lim_{x \to 0} ~ f(x)g(x)=0
$
This is similar. Try it.

3. wow, that looks so simple..

so does this look about right?

$
\forall x \in R\ g(x)f(x) \leq Bf(x).$

$So\ lim_{x \to 0} ~ Bf(x)=0$
$
Thus,\ lim_{x \to 0} ~ f(x)g(x)=0$

4. Originally Posted by Pisces
wow, that looks so simple..

so does this look about right?

$
\forall x \in R\ g(x)f(x) \leq Bf(x).$

$So\ lim_{x \to 0} ~ Bf(x)=0$
$
Thus,\ lim_{x \to 0} ~ f(x)g(x)=0$
not exactly. with that inequality, it would be best for you to use the squeeze theorem here, which you did not

as TPH did, consider $|g(x)f(x)|$

5. I'm confused.

Could you give me a hint?

edit: im not even sure i understand how he did the first one.

6. Originally Posted by Pisces
I'm confused.

Could you give me a hint?

edit: im not even sure i understand how he did the first one.
$|g(x)f(x)| \le |Bf(x)| = B|f(x)|$

now take the limit of that system

you should look up the squeeze theorem

7. I see the theorem, and I understand h(x)<=f(x)<=g(x) and if the limit of h(x)=0 and limit of g(x)=0 then the limit of f(x)=0..

but all this other stuff isn't really clicking.

edit: plz help I am really confused.

is that not what I did?

$
lim_{x \to 0} ~ Bf(x)=0$

$so\ lim_{x \to 0} ~ f(x)g(x)=0
$

8. Originally Posted by Pisces
2. $
Suppose\ that\ \lim_{x \to 0} ~ f(x)=0\ and\ that\ \forall x \in R\ g(x) \leq B.$

$Show \lim_{x \to 0} ~ f(x)g(x)=0
$
I think this is wrong unless we have for all real number $x$, ${\color{red}|}g(x){\color{red}|}\leq B$.