# 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

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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 [(x^2-x-2)/(x-2)]

=> lim x--> 2 [(x^2-2x + x - 2)/(x-2)]

=> lim x--> 2 [(x(x - 2) + 1(x - 2))/(x-2)]

=> lim x--> 2 [(x + 1)(x - 2)/(x-2)]

=> lim x--> 2 [(x + 1)]

=> 3

We see that the function is defined at f(2) as 1 but for lim x--> 2, the function gives 3.

**Therefore the given function is discontinuous at x = 2.**

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.**