# 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!