# simple question about epsilon and estimation turing machines

i am getting really confused by it. i got to a point i had to calculate the lim when $$n \rightarrow \infty$$ for an optimization problem, and i got to the point that i had to calculate a fairly simple limit: $$lim_{n \rightarrow \infty} {3-\frac{7}{n}}$$.

now i used $$3 - \epsilon$$ and i am trying to show that there can't be any $$\epsilon>0$$ so that the estimation of the algorithm is $$3-\epsilon$$, because there exists a "bigger estimation" - and this is the part i am not sure about, what is the correct direction of the inequality? $$3-\frac{7}{n} > 3 - \epsilon$$ or the opposite? i am trying to show that the estimation ration is close to 3.

i think that what i wrote is the correct way, but not sure. would appreciate knowing what is correct in this case. thanks.

Suppose there exists $$k>0$$ such that $$\lim_{n\to\infty} \left(3-\frac7n\right)=3-k$$.
By the definition of a limit, $$\lim_{n\to\infty} f(n)=L\Leftrightarrow \forall\epsilon>0,\exists N>0,n>N\implies |f(n)-L|<\epsilon$$
Here, we need to check: $$\lim_{n\to\infty} \left(3-\frac7n\right)=3-k\Leftrightarrow \forall\epsilon>0,\exists N>0,n>N\implies \left|k-\frac7n\right|<\epsilon$$
Now, disproving this just requires a suitable choice of $$\epsilon$$. Consider $$\epsilon=\frac k2$$. We have for all $$n>\frac{14}{k}$$, $$\left|k-\frac7n\right|>\frac k2$$. So, there does not exist any $$N>0$$ satisfying the condition, and as a result, no such $$k>0$$ can exist.