# Proving that 1/(x+1) is continuous at x=0 using epsilon-delta def.

## Main Question or Discussion Point

As the thread title. The question is actually:
"Given p > 0, find d so that |x-0| < d implies |f(x)-1| < p and hence deduce that f(x) approaches 1 as x approaches 0."

My problem is that, when x is near the point x=0, x can be positive and negative, I don't know how to get my delta value because |f(x) -1| = |-x/(1+x)| can't be got into the |x-0| form without worrying about the signs of everything. It's confusing.

Any help?

Note that |x-0|=|x| and |-x|=|x|. So,
$$|f(x)-1| = \left| \frac{-x}{x+1} \right| = \left| \frac{x}{x+1} \right|$$
Now, can you pick your d so that if |x|<d, then |f(x)-1|<p? It helps to allow your d to rely on p.

Note that |x-0|=|x| and |-x|=|x|. So,
$$|f(x)-1| = \left| \frac{-x}{x+1} \right| = \left| \frac{x}{x+1} \right|$$
Now, can you pick your d so that if |x|<d, then |f(x)-1|<p? It helps to allow your d to rely on p.
Well when I get to $$|\frac{x}{x+1}|$$ and try to find something it's less than I can't get anywhere. So I can't say $$|\frac{x}{x+1}| <|x|$$ because if x is negative then the denominator is <1. So I don't know how to simplify it to get a value out for delta. Could you give me a bigger hint? Thanks.

Hurkyl
Staff Emeritus
Gold Member
if x is negative then the denominator is <1
Okay, the denominator is a problem. Can you choose $\delta$ in some way that lets you control what range of values the denominator takes?

Office_Shredder
Staff Emeritus
Gold Member
$$|\frac{x}{x+1}| = \frac{1}{|1+\frac{1}{x}|}$$

Now use what you know about algebra of limits

Hurkyl
Staff Emeritus
Gold Member
Now use what you know about algebra of limits
Just to elaborate on this -- while the problem asks you to make an epsilon-delta proof, it does not forbid you from using your knowledge of the algebra of limits to figure out how such a proof should look. You can often directly translate a proof by algebra of limits into an epsilon-delta proof.

By the way, you learned a lot about solving inequalities in precalculus, didn't you? Along with how to solve equations involving absolute values? Have you considered using any of that knowledge?

Okay, the denominator is a problem. Can you choose $\delta$ in some way that lets you control what range of values the denominator takes?
All I can think of is to make x>0, but then I think that'll go down to taking a limit from the RHS. Are you saying I should take the LHS and RHS limits to get this out?

As for the inequalities knowledge: I guess I have learnt quite a bit but I have no idea how else I can approach this problem than the way I'm doing now. Sorry :(

Hurkyl
Staff Emeritus
Gold Member
As for the inequalities knowledge: I guess I have learnt quite a bit but I have no idea how else I can approach this problem than the way I'm doing now. Sorry :(
You have an inequation:
|f(x) - 1| < p​
you have the knowledge to solve that equation for x.

Ok, so if I solve the inequality, I get |x| < p/(1-p). Is that right?

just curious, by why work so hard to prove its continuous there?
the right and left handed limits are the same, and the point exists, so its continuous. why does one need epsilon-delta?

Hurkyl
Staff Emeritus