What is limx-->0 f(x) when lim x-->0 g(x)=0?

[lep11]

Homework Statement

Let |f(x)|≤g(x) for all x∈ M_f∩M_g.
What is lim_x-->0 f(x) when lim _x-->0 g(x)=0?
What is lim_x-->0 f(x) when lim _x-->0 g(x)=3?

The Attempt at a Solution

Well, given that|f(x)|≤g(x), lim _x-->0 g(x)=0 intuitively implies to me that | lim_x-->0 f(x) |≤0
therefore | lim_x-->0 f(x) |=0 --> lim_x-->0 f(x)=0

AND when lim _x-->0 g(x)=3

|lim_x-->0 f(x)|≤3 ⇔ -3 ≤lim_x-->0 f(x)≤3

Is my reasoning correct?

 

[pasmith]

Homework Statement

Let |f(x)|≤g(x) for all x∈ M_f∩M_g.
What is lim_x-->0 f(x) when lim _x-->0 g(x)=0?
What is lim_x-->0 f(x) when lim _x-->0 g(x)=3?

The Attempt at a Solution

Well, given that|f(x)|≤g(x), lim _x-->0 g(x)=0 intuitively implies to me that | lim_x-->0 f(x) |≤0

Can you turn your intuition into a rigorous proof that $\lim_{x \to 0} f(x) = 0$ in this case?

AND when lim _x-->0 g(x)=3

|lim_x-->0 f(x)|≤3 ⇔ -3 ≤lim_x-->0 f(x)≤3

Is my reasoning correct?

Must $\lim_{x \to 0} f(x)$ necessarily exist when $\lim_{x \to 0} g(x) > 0$? Consider the case $$f(x) = \begin{cases} -1 & x < 0, \\ 0 & x = 0, \\ 1 & x > 0, \end{cases}\qquad g(x) = 3.$$

 

[lep11]

Can you turn your intuition into a rigorous proof that $\lim_{x \to 0} f(x) = 0$ in this case?

How?

Must $\lim_{x \to 0} f(x)$ necessarily exist when $\lim_{x \to 0} g(x) > 0$? Consider the case $$f(x) = \begin{cases} -1 & x < 0, \\ 0 & x = 0, \\ 1 & x > 0, \end{cases}\qquad g(x) = 3.$$

No. So in that case we cannot say anythin about $\lim_{x \to 0} f(x)$ since it won't necessarily even exist?

However if it exists then -3 ≤lim_x-->0 f(x)≤3

 

[lep11]

ahh there is a small typo in the problem statement.. it should be x approaches a, not 0

 

[lep11]

I used the squeeze theorem to prove that lim_x→0f(x)=0

 

[Mark44]

ahh there is a small typo in the problem statement.. it should be x approaches a, not 0

I used the squeeze theorem to prove that lim_x→0f(x)=0

Isn't it the limit as x approaches a?So for the second part, if $|f(x)| \le g(x)$ and $\lim_{x \to a} g(x) = 3$, is it necessarily true that $\lim_{x \to a} f(x) = 0$?

 

[lep11]

Isn't it the limit as x approaches a?

Yes it is, I'm sorry

So for the second part, if $|f(x)| \le g(x)$ and $\lim_{x \to a} g(x) = 3$, is it necessarily true that $\lim_{x \to a} f(x) = 0$?

No, because limx→a f(x) won't necessarily exist when lim x-->a g(x)=3.

 

[Mark44]

I used the squeeze theorem to prove that lim_x→0f(x)=0

So for the second part, if $|f(x)| \le g(x)$ and $\lim_{x \to a} g(x) = 3$, is it necessarily true that $\lim_{x \to a} f(x) = 0$?

No, because limx→a f(x) won't necessarily exist when lim x-->a g(x)=3.

OK, I wasn't sure whether you were using the squeeze theorem for the second part.