Proving Continuity of a Rational Function

AI Thread Summary
The discussion centers on proving the continuity of the rational function (2x-1)/(x^2+1) for all x in the real numbers. Participants reference the definition of continuity and the Algebraic Continuity Theorem, emphasizing that both the numerator and denominator functions are continuous, with the denominator never equaling zero. They confirm that the function meets the three conditions for continuity: being defined at a point, having a limit as x approaches that point, and the limit equating to the function's value at that point. Additionally, there is a mention of proving continuity for other functions, such as sin(2x+3) and ln(x-2), with specific conditions noted for the latter. The conversation highlights the importance of understanding and applying continuity definitions and theorems in rational functions.
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Homework Statement



Prove that the function:

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

is continuous.

Homework 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.

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 ?
 
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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?
 
Yes, that's what I thought.
But how will I prove for f(x)=x/(x+1), x Є R \ {-1} ?
 
Do you know (or are allowed to use) the Algebraic Continuity Theorem?
 
Дьявол said:
Yes, that's what I thought.
But how will I prove for f(x)=x/(x+1), x Є R \ {-1} ?

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.
 
VeeEight said:
Do you know (or are allowed to use) the Algebraic Continuity Theorem?
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.
Mark44 said:
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.
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

Thanks in advance.
 
Last edited:
Yes, except for ln(x - 2), it must be that x > 2.
 
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