Question 1163182
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<pre>

In vicinity of the point x= -2,  


    x^2 - x - 6 = (x-3)*(x+2),


so, after dividing by (x+2)


    f(x) = {{{((x-3)*(x+2))/(x+2)}}} = x-3  at  x < -2.


So, the limit of f(x) at x---> -2 from the left  is equal to -5.


From the other side,  f(-2) = -5, by the definition of the function f(x)  (second line).


Also, the limit f(x) at x ---> -2 from the right is equal to -5.


So, the function f(x) is CONTINUOUS at x= -2.




Next, at x= 3, the function f(x) is not formally defined, so we even can not discuss the question if it is continuous at this point.


It is NOT.
</pre>

Solved, answered and explained.



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An important post-solution note.


<pre>
    The way how the function f(x) is defined in the third line, 


        f(x) = { (x^2-9)/(x-3)  , if x≥3


    is a  <U>HUGE mistake</U>.   It is impossible to define a function in this way, since division by  0  at x= 3 is STRICTLY PROHIBITED.


    The correct form of this third line should be THIS


        f(x) = { (x^2-9)/(x-3)  , if x > 3.


    At x = 3, the function f(x) is UNDEFINED.
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