## anonymous one year ago Guess the value of the limit:

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1. anonymous

2. misty1212

HI!!

3. misty1212

make i guess, i bet you get it on the first try what is $$1.999$$ close to?

4. misty1212

hint, same number that $$2.0001$$ is close to

5. amistre64

... imaginary numbers are soo complicated

6. Owlcoffee

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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