I Limit of a product with bounded function

[Austin Chang]

Q's Let f,g ℝ→ℝ. Suppose that g is bounded. This means that its image is bounded or in other words there exists a positive real number B s.t. |g(x)| ≤ B ∀ x. Prove that if lim x→c f(x) = 0, then lim x→c f(x)g(x) = 0.
Work.
See the picture.
I am really confused I can't seem to understand the idea or the concept behind this. Can anyone explain in terms of like actual terms for example f(x) = x and g(x)= 1/(1+x^2) and parse through the idea with me? I kind of understand it but what is really messing me up is the min{,} I really need to understand that idea and that will really help me understand the idea as a whole.

Thanks

 

[fresh_42]

I cannot see your image very well and don't want to spend more time to make it readable than it took you to type it in here.

The basic idea is, that when $f(x)$ tends to zero, then $g(x)$ being bounded cannot outperform $f(x)$.
Look at an example in which it is not the case:
If $f(x)=x$ then $\lim_{x \rightarrow 0} f(x) = 0$ and $g(x) := x^{-2}$ lead to $\lim_{x \rightarrow 0} f(x)g(x) = \lim_{x \rightarrow 0} x^{-1} = \infty$. Here $g$ outperforms the limitation given by $f$.

However, if $g$ is bounded, i.e. $-B \leq g(x) \leq B$, then $-B\cdot |f(x)| \leq |f(x)| \leq B\cdot |f(x)|$. Now take the limit $x \rightarrow c$ of these inequations.

 

[Austin Chang]

I cannot see your image very well and don't want to spend more time to make it readable than it took you to type it in here.

The basic idea is, that when $f(x)$ tends to zero, then $g(x)$ being bounded cannot outperform $f(x)$.
Look at an example in which it is not the case:
If $f(x)=x$ then $\lim_{x \rightarrow 0} f(x) = 0$ and $g(x) := x^{-2}$ lead to $\lim_{x \rightarrow 0} f(x)g(x) = \lim_{x \rightarrow 0} x^{-1} = \infty$. Here $g$ outperforms the limitation given by $f$.

However, if $g$ is bounded, i.e. $-B \leq g(x) \leq B$, then $-B\cdot |f(x)| \leq |f(x)| \leq B\cdot |f(x)|$. Now take the limit $x \rightarrow c$ of these inequations.

how about if B < 1?

 

[fresh_42]

how about if B < 1?

How would this affect $\lim_{x \rightarrow c} (-B\cdot |f(x)|) \leq \lim_{x \rightarrow c} |f(x)| \leq \lim_{x \rightarrow c}(B\cdot |f(x)|)\;$, then $\lim_{x \rightarrow c} |f(x)g(x)|\;$ and at last $\lim_{x \rightarrow c} f(x)g(x)\;$?

 

[Austin Chang]

Would it not cause
limx→c(B⋅|f(x)|) to be less than limx→c|f(x)|

 

[fresh_42]

Oh sorry, I've made a typing error and repeated it without looking at it again.
It should have been
$$-B \leq g(x) \leq B$$
$$\Rightarrow -B\cdot |f(x)| \leq |f(x)|\cdot g(x) \leq B\cdot |f(x)|$$
$$\Rightarrow \lim_{x \rightarrow c} (|f(x)|\cdot g(x)) = 0$$
$$\Rightarrow \lim_{x \rightarrow c} |f(x)\cdot g(x)| = 0$$
$$\Rightarrow \lim_{x \rightarrow c} (f(x)\cdot g(x)) = 0$$