# Thread: Limits (With a specified range?)

1. ## Limits (With a specified range?)

Suppose 0≤ f(x) ≤1 for all x, find

lim xf(x)
x→0.

This is exactly how the question is presented in my manual and I am not sure if it is trying to say that the function is x⁴ or something else. Also, I am not sure what relevance the first part of the question has to finding the limit. I believe it is representing the range, although I don't know what that means as far as limits go. Thank-you for your help.

2. ## Re: Limits (With a specified range?)

The function $f(x)$ can't be $x^4$, because $x^4$ isn't surely bounded by $[0,1]$ for all $x$, but if you rewrite:
$\lim_{x\to 0} x^4\cdot f(x)=\lim_{x\to 0}x^4\cdot \lim_{x\to 0} f(x)=...$ and you know $0\leq f(0)\leq 1$ then the limit is ...

3. ## Re: Limits (With a specified range?)

I am still a little unclear as to the part which states:

0≤ f(x) ≤1 for all x.

Does this represent the range of f(x)?

I understand that if the limit of
lim x⁴
x→0

is 0 that multiplying this by the limit of f(x) will yield a limit of 0. I am just wondering about the relevance of the range stated and whether that should affect my solution. Thank-you.

4. ## Re: Limits (With a specified range?)

Originally Posted by Pewter12
0≤ f(x) ≤1 for all x.
Does this represent the range of f(x)?
I understand that if the limit of
lim x⁴x→0 is 0 that multiplying this by the limit of f(x) will yield a limit of 0.
Actually this is the relevance of that condition.
$0\le f(x)\le 1$ implies that $0\le x^4f(x)\le x^4$.
From that it follows at once that the limit is zero.

We do not know that $\lim _{x \to 0} f(x) = f(0)$

5. ## Re: Limits (With a specified range?)

I'm sorry, what I meant was that the product of

lim x⁴ x lim f(x)
x→0 x→0

would be 0 because

lim x⁴
x→0 = 0

Is there a particular rule telling us that
0≤ f(x) ≤1

implies also that

0≤ x⁴f(x) ≤1?

6. ## Re: Limits (With a specified range?)

Originally Posted by Pewter12
Is there a particular rule telling us that
0≤ f(x) ≤1
implies also that 0≤ x⁴f(x) ≤1?
Well I did not say that it did: $0\le x^4f(x)\le x^4$.
However, if $|x|<1$ then $x^4<1$.
So in the limit process that does in fact follow.

But my point in post #4 is that you use the squeeze play.