Proving Limit Exists: x-2 of f(x)=2

[rb120134]

Homework Statement

Prove that delta>0 exists such that f(x)>1 using delta epsilon definition

Relevant Equations

[x-a]<delta such that f(x)>1

Given is the following: lim x-2 of f(x)=2 prove (using delta, epsilon definition of a limit) that a delta exists so that when [x]<delta then f(x)>1
I came up with when [x-a]<delta (f(a)-epsilon<f(x)< f(a) + epsilon) so f(a)-epsilon>1 so epsilon<f(a) -1 but I don't know how to prove this or how to answer this question?

 

[PeroK]

Perhaps someone else can make sense of that, but that is just a muddle to me. What is the question exactly?

 

[Mark44]

Homework Statement:: Prove that delta>0 exists such that f(x)>1 using delta epsilon definition
Relevant Equations:: [x-a]<delta such that f(x)>1

Given is the following: lim x-2 of f(x)=2 prove (using delta, epsilon definition of a limit) that a delta exists so that when [x]<delta then f(x)>1
I came up with when [x-a]<delta (f(a)-epsilon<f(x)< f(a) + epsilon) so f(a)-epsilon>1 so epsilon<f(a) -1 but I don't know how to prove this or how to answer this question?

I agree with PeroK's assessment.
Is it given that f(x) = 2 is a constant function, or are you trying to prove this limit?
$$\lim_{x \to 2} f(x) = 2$$

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