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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.

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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.

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