The given limit indicates that there is a vertical asymptote at x = 2.
The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator. When x tends to these values laterally, the limit of f(x) goes to infinity.
In this problem, the function is:
[tex]f(x) = \frac{x^2 - 4x - 12}{x - 2}[/tex]
The vertical asymptote is:
x - 2 = 0 -> x = 2.
Hence:
[tex]\lim_{x \rightarrow 2^+} f(x)[/tex] and [tex]\lim_{x \rightarrow 2^-} f(x)[/tex] are either positive infinity or negative infinity.
More can be learned about asymptotes at https://brainly.com/question/16948935
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