Homework Help: Continuity function

1. Sep 20, 2014

Unusualskill

(a) State precisely the definition of: a function f is continuous at a point
a ∈ R.

(b) At which points x ∈R is the function:
f(x) = sin(1/x)continuous?
You may assume that g(x) = 1=x is continuous on its domain, and
h(x) = sin(x) is continuous on its domain.

(c) Let f and g be functions such that:
1. g is continuous at 0, and g(0) = 0.
2. For all x∈ R, lf(x)l <= lg(x)l.
Use the
ϵ− δ definition of limit to show that f is continuous at 0.

I did part (a) and (b) but i don understand how to do part(c).Can any1 provide guidance on it?thanks

2. Sep 20, 2014

slider142

First, write down precisely what the definition of continuity requires in order for an arbitrary function f to be continuous at 0. Then, if it is not already written in terms of ϵ− δ, translate the limit into the exact ϵ− δ definition of a limit. You should now have an ϵ− δ definition of what it would mean for an arbitrary function f to be continuous at 0. If you are unsure about any of this, write down exactly what you did for these parts here on the forum.
Now, let f be any function that satisfies requirement (2) of part (c). Do you see anything in your definition that can now be simplified in this special case?

3. Sep 20, 2014

Unusualskill

What I did:
For f to be continuous at 0, lim f(x) as x approaches to 0 =f(0).
Thus, lim f(x) at x approaches to 0 exists and equal to f(0). If for every epsilon, there exists a delta such that if x satisfies
0<lxl<delta then lf(x)-f(0)l<epsilon
Can guide me on how to continue ?thank you

4. Sep 20, 2014

slider142

Great. This is the statement that we would like to be true; that is, we want to prove that this statement is true for the particular function f described by precepts (1) and (2).
So far, our definitions only involve one function: f. Precept (2) tells us that the particular function we want to consider must have the property that "$|f(x)| \leq |g(x)|$ for all values of x". We would like to somehow replace all occurrences of f with g, since we know that g is continuous at 0 (precept (1)).
Hmm. We know that we would like $|f(x) - f(0)| < \epsilon$. We would like to claim that there is a value of $\delta$ that makes this inequality true. But wait. We do have a value of $\delta$ that makes g continuous at 0. Let us write out exactly what it means, in $\epsilon-\delta$ form, for g to be continuous at 0, given that g(0) = 0.
Our plan is to use this inequality, along with the inequality relating f and g, to claim that there is a value of $\delta$ for which $|f(x) - f(0)| < \epsilon$.

Last edited: Sep 20, 2014
5. Sep 20, 2014

Unusualskill

Thanks for your guidance. Now, I reached lf(x)l<=lg(x)l<epsilon >>>>lf(x)l<epsilon... How should I continue?

6. Sep 20, 2014

slider142

That's perfect. Now we know that for every value of $\epsilon > 0$, there is some value of $\delta > 0$ such that, for all x, if $x < \delta$, then $|f(x)| < \epsilon$. We would just like to change that last phrase to $|f(x) - f(0)| < \epsilon$.
Remember that $|f(x)| \leq |g(x)|$ for all values of x. What does this imply about f(0) ?

7. Sep 20, 2014

Unusualskill

Can you tell me what is it or some hints?

8. Sep 20, 2014

slider142

Since $|f(x)| \leq |g(x)|$ for all values of x, we must have $|f(0)| \leq |g(0)|$. Since g(0) = 0, what does this imply about f(0) ?

9. Sep 21, 2014

Unusualskill

Thanks alot. It implies that f(0) must be 0 also...!thanks!