I was trying to prove that the function: [math]f(x)=\frac{x+1}{x^2+1}[/math] is continuous over the real Nos

 

And in considering |f(x)-f(a)| I come up with the inequality:[math]|f(x)-f(a)|\leq\frac{|x-a|(|ax|+|a|+|x|+1)}{(x^2+1)(a^2+1)}[/math]

 

And in taking values of x near a ,i.e |a-x|<1 i come up with the inequality:[math]\frac{|x-a|(|ax|+|a|+|x|+1)}{(x^2+1)(a^2+1)}\leq\frac{|a-x|(|a|^2+3|a|+2)}{(x^2+1)(a^2+1)}[/math]

 

And here i stop

 

Any ideas how to get rid of x^2+1 in the denominator??

I was trying to prove that the function: [math]f(x)=\frac{x+1}{x^2+1}[/math] is continuous over the real Nos

 

And in considering |f(x)-f(a)| I come up with the inequality:[math]|f(x)-f(a)|\leq\frac{|x-a|(|ax|+|a|+|x|+1)}{(x^2+1)(a^2+1)}[/math]

 

And in taking values of x near a ,i.e |a-x|<1 i come up with the inequality:[math]\frac{|x-a|(|ax|+|a|+|x|+1)}{(x^2+1)(a^2+1)}\leq\frac{|a-x|(|a|^2+3|a|+2)}{(x^2+1)(a^2+1)}[/math]

 

And here i stop

 

Any ideas how to get rid of x^2+1 in the denominator??

 

Probably the easiest and most instructive way to proceed is to prove a series of lemmas

 

1. Inversion, the function f defined by f(x)= 1/x is continous except at 0.

 

2. The fulnction f defined by f(x) = x is continuous everywhere.

 

2. If f and g are continuous so is the product [math] f \times g [/math]

 

3. If f and g are continyous so is composition [math] f \circ g[/math]

1+x^2 has a minimum at x = 0, so 1/(1+x^2) has a max at x = 0 with the value 1. Net result: your last expression is bounded by |a-x|(a^2 + 3|a| + 2).

Probably the easiest and most instructive way to proceed is to prove a series of lemmas

 

1. Inversion, the function f defined by f(x)= 1/x is continous except at 0.

 

2. The fulnction f defined by f(x) = x is continuous everywhere.

 

2. If f and g are continuous so is the product [math] f \times g [/math]

 

3. If f and g are continyous so is composition [math] f \circ g[/math]

 

Sorry ,i did not mentioned it in my original post ,but i meant to solve the problem using the epsilon - delta definition

1+x^2 has a minimum at x = 0, so 1/(1+x^2) has a max at x = 0 with the value 1. Net result: your last expression is bounded by |a-x|(a^2 + 3|a| + 2).

 

 

you mean that :[math]\delta[/math]=min{[math] 1,\frac{\epsilon}{(a^2+3|a|+2)}[/math]} ??

Sorry ,i did not mentioned it in my original post ,but i meant to solve the problem using the epsilon - delta definition

 

One can, and should, use the "epsilon-delta definition" to prove each of the lemmas that I suggested.

 

Breaking the problem down to the fundamental parts can, and often does, illuminate what is important and makes the proofs easier to formulate and understand. You can go after the problem directly, but you risk losing the forest for the trees amidst some messy algebra.

you mean that :[math]\delta[/math]=min{[math] 1,\frac{\epsilon}{(a^2+3|a|+2)}[/math]} ??

I guess so, but I don't see any δ or ε in your description.

I guess so, but I don't see any δ or ε in your description.

 

As i said in my other post ,i forgot to mention it in my OP

As i said in my other post ,i forgot to mention it in my OP

 

Your other post says you want to use epsilon-delta definition, but no specifics, so my answer is still I guess so.