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

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.0make i guess, i bet you get it on the first try what is \(1.999\) close to?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0hint, same number that \(2.0001\) is close to

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0... imaginary numbers are soo complicated

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0A 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: \[\lim_{x \rightarrow 2}\frac{ x^22x }{ x^2x2 }\] 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 xvalues very close to x=2: \(f(x)=2.001\), \(x \rightarrow 2^{+} \) \(f(x)=1.999\), \(x \rightarrow 2^{} \)
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