# [prove] Continuous function

1. Nov 27, 2008

### Дьявол

1. The problem statement, all variables and given/known data

Prove that the function:

$$\frac{2x-1}{x^2+1}, x \in \mathbb{R}$$

is continuous.

2. Relevant equations
Definition 1.

The function y=f(x) satisfied by the set Df is continuous in the point x=a only if:

10 f(x) is defined in the point x=a i.e. $a \in D_f$

20 there is bound $$\lim_{x \rightarrow a}f(x)$$

30 $$\lim_{x \rightarrow a}f(x)=f(a)$$

Theorem 1.
If the functions y=f(x) and y=g(x) are continuous in the point x=a Є Df ∩ Dg, then in the point x=a these functions are continuous:
y=f(x)+g(x), y=f(x)g(x) and y=f(x)/g(x), if g(a) ≠ 0.

3. The attempt at a solution

I tried using the definition 1.

But also this function is composition of two functions f(x) and g(x), so can I use the fact that f(x)=2x-1 and g(x)=x2+1 are continuous, and y=f(x)/g(x), g(a) ≠ 0 since x2+1 ≠ 0 ?

2. Nov 27, 2008

### VeeEight

Let f(x) = 2x - 1 and g(x) = x^2 + 1. Are f(x) and g(x) continuous functions? Is f(x)/g(x) continuous on it's domain?

3. Nov 27, 2008

### Дьявол

Yes, that's what I thought.
But how will I prove for f(x)=x/(x+1), x Є R \ {-1} ?

4. Nov 27, 2008

### VeeEight

Do you know (or are allowed to use) the Algebraic Continuity Theorem?

5. Nov 27, 2008

### Staff: Mentor

You can show that f(x) = x/(x + 1) satisfies all three of the conditions you listed in your first post. I.e., a) that f is defined at a (where a != -1, which is not in R \ {-1}), b) lim f(x) as x approaches a exists, and c) lim f(x) = f(a), as x approaches a.

6. Nov 28, 2008

### Дьявол

VeeEight could you please specify on what theorem do you mean? I am supposed to use the definition 1. or theorem 1. in the first post.
10 f(a)= a/(a+1)

20 $$\lim_{x \rightarrow a}f(x)=\lim_{x \rightarrow a}\frac{x}{x+1}=\frac{\lim_{x \rightarrow a}(x)}{\lim_{x \rightarrow a}(x+1)}=\frac{a}{a+1}$$

30 $$\lim_{x \rightarrow a}f(x)=\frac{a}{a+1}=f(a)$$

Should I prove the other tasks like this?

Because I got:

f(x)=sin(2x+3), x Є R

and

f(x)=ln(x-2), x Є R

Last edited: Nov 28, 2008
7. Nov 28, 2008

### Staff: Mentor

Yes, except for ln(x - 2), it must be that x > 2.