anonymous
  • anonymous
Guess the value of the limit:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
make i guess, i bet you get it on the first try what is \(1.999\) close to?

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misty1212
  • misty1212
hint, same number that \(2.0001\) is close to
amistre64
  • amistre64
... imaginary numbers are soo complicated
Owlcoffee
  • 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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