Question

Guess the value of the limit:

Answer 1

HI!!

Answer 2

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

Answer 3

hint, same number that \(2.0001\) is close to

Answer 4

... imaginary numbers are soo complicated

Answer 5

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:
\(f(x)=2.001\), \(x \rightarrow 2^{+} \)

\(f(x)=1.999\), \(x \rightarrow 2^{-} \)