Where is the function f(x) discontinuous? f(x) = (x^2-x-2)/(x-2), if x is not 2 f(x) = 1, if x=2
We have the function:
f(x) = (x^2-x-2)/(x-2), if x is not 2
f(x) = 1, if x=2
At x = 2, f(x) = 1
For lim x-->2 [f(x)]
=> lim x--> 2...
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For a function to be continuous, the values of the lateral limits have to be equal and they are equal to the value of function, if we'll let x to be equal to the value of accumulation point.
We'll calculate the lateral limits, if x approaches to 2 from the left and from the right:
lim (x^2-x-2)/(x-2) = lim (x-2)(x+1)/(x-2) = lim (x+1) = 2+1 = 3
Now, we'll evaluate the limit if x = 2.
According to enunciation, f(2) = 1.
We notice that the limit of the function, if x approaches to 2, is not equal to the value of the function f(2).
The function is discontinuous at x = 2.