Guess the value of the limit:

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Guess the value of the limit:

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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HI!!
make i guess, i bet you get it on the first try what is \(1.999\) close to?

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hint, same number that \(2.0001\) is close to
... imaginary numbers are soo complicated
A limit is defined the following way: \[\lim_{x \rightarrow a}f(x)=b <=> f/e E(b, \epsilon) \exists E ^{*}(a, \delta) / \forall x \in E ^{*}(a,\delta) : f(x) \in E(b, \epsilon)\] This means, in essence that we consider an "enviroment" whose center is x=a and the radius is delta. Now, when we say an "enviroment" we consider all the values of "x" that are infintely close to x=a but never actually take the value x=a. With that in mind, let's return to the excercise in question: \[\lim_{x \rightarrow 2}\frac{ x^2-2x }{ x^2-x-2 }\] I will suppose the values below are very proximate values to x=2, and if you recall, that is an enviroment whose center is "x=2". take a look at the final values given, they respresent the values of the function as we move all the x-values very close to x=2: \(f(x)=2.001\), \(x \rightarrow 2^{+} \) \(f(x)=1.999\), \(x \rightarrow 2^{-} \)

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