Proof that a function is continuous on its domain

Whistlekins
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Homework Statement



We have f(x) = \frac{x^{2}+x-2}{x-1}+cos(x) , x\in\mathbb{R}\setminus \{1\} and wish to prove that it is continuous on its domain.

Homework Equations



The delta-epsilon definition of the continuity of a function.

The Attempt at a Solution



I've managed to reduce |f(x) - f(x_0)|to|x-x_0| + |cos(x) - cos(x_0)| < \delta + |cos(x) - cos(x_0)|
I'm not too sure where to go from there or even if I'm on the right track. Any insight would be greatly appreciated.
 
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Whistlekins said:

Homework Statement



We have f(x) = \frac{x^{2}+x-2}{x-1}+cos(x) , x\in\mathbb{R}\setminus \{1\} and wish to prove that it is continuous on its domain.

Homework Equations



The delta-epsilon definition of the continuity of a function.

The Attempt at a Solution



I've managed to reduce |f(x) - f(x_0)|to|x-x_0| + |cos(x) - cos(x_0)| < \delta + |cos(x) - cos(x_0)|
I'm not too sure where to go from there or even if I'm on the right track. Any insight would be greatly appreciated.

Factoring x^2+x-2 would be a great first step.
 
Dick said:
Factoring x^2+x-2 would be a great first step.

I have, that's how I arrived at the reduced expression. The questions is where to go from there. I could always use the property that if f and g are continuous at a point x_0\in \mathbb{A} then f+g is continuous at x_0. But I don't know how to prove that cos(x) is continuous on the domain.
 
Whistlekins said:
I have, that's how I arrived at the reduced expression. The questions is where to go from there. I could always use the property that if f and g are continuous at a point x_0\in \mathbb{A} then f+g is continuous at x_0. But I don't know how to prove that cos(x) is continuous on the domain.

Yes, that you did. One way to prove cos is continuous is to use the trig identity, cos(u)-cos(v)=(-2)sin((u+v)/2)*sin((u-v)/2).
 
Dick said:
Yes, that you did. One way to prove cos is continuous is to use the trig identity, cos(u)-cos(v)=(-2)sin((u+v)/2)*sin((u-v)/2).

Ahh I overlooked that, thanks! I think I've got it now.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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