Can the Limit of a Function Exist Despite Contradictory Values?

[Dank2]

Homework Statement

Proof that the limit of the function below doesn't exists.
lim_x-->1 1/(x-1)

Homework Equations

The Attempt at a Solution

Lets assume that limit L exists.
So if (1) 0< |x-1| < δ then (2) |1/(x-1) - L| < ε

at the book they gave an example by giving a value to ε.
put ε = 1. then showing a contradiction by giving two δ values to x.

but now I am thinking about what values can i put that satisfy (1) that for them |1/(x-1) - L| < 1 doesn't hold.

 

[Dank2]

Maybe if ε = L/2

then if i put x₁= δ/2+1, x₂ = -δ/2 +1 both of them satisfy (1)
then i get for (2)
|δ/2 - L| < L/2, and |-δ/2 - L| < L/2 δ>0
and we can see that |-δ/2 - L| < L/2 doesn't hold. and that is contradiction, and therefore the limit doesn't exists.

 

[Dank2]

now i see i haven't shown that L ≠ 0, because ε > 0.ε = δ/2, looks like it would work now. since δ >0.

|δ/2 - L| < δ/2, and |-δ/2 - L| < δ/2

and in all cases of L, L>0, L<0. L=0 there are contradictions.
and therefore there is no limit.

 

[Mark44]

Homework Statement

Proof that the limit of the function below doesn't exists.
lim_x-->1 1/(x-1)

Homework Equations

The Attempt at a Solution

Lets assume that limit L exists.
So if (1) 0< |x-1| < δ then (2) |1/(x-1) - L| < ε

at the book they gave an example by giving a value to ε.
put ε = 1. then showing a contradiction by giving two δ values to x.

but now I am thinking about what values can i put that satisfy (1) that for them |1/(x-1) - L| < 1 doesn't hold.

Instead of a proof by contradiction, why don't you try proving this directly? Looking at the graph of f(x) = 1/(x - 1), it's clear that the limit doesn't exist (in any sense), because $\lim_{x \to 1^-}\frac 1 {x - 1} = -\infty$ while $\lim_{x \to 1^+}\frac 1 {x - 1} = \infty$

If you can use the definition of the limit to prove each of these one-sided limits, that should do the job.

 

[Dank2]

Instead of a proof by contradiction, why don't you try proving this directly? Looking at the graph of f(x) = 1/(x - 1), it's clear that the limit doesn't exist (in any sense), because $\lim_{x \to 1^-}\frac 1 {x - 1} = -\infty$ while $\lim_{x \to 1^+}\frac 1 {x - 1} = \infty$

If you can use the definition of the limit to prove each of these one-sided limits, that should do the job.

Sorry, yes you are right, i can do that too.