Proof that a function is continuous on its domain

[Whistlekins]

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.

 

[Dick]

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.

 

[Whistlekins]

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.

 

[Dick]

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).

 

[Whistlekins]

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.