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\[f(x)= \frac{ 2x-4 }{ 3x-2 } at x=2\]

remember the definition: f is continous at x=a if
\[ \large \lim_{x\to a}f(x)=f(a) \]

yes it is continuous..

it is defined at x=2 right

f(2)=0
\[\large \lim_{x\to 2}f(x)=0=f(2)\]

the func. is discontinuous when denominator is 0

2x-6=0
x=3
so, for x=3 the func. is not continuous

and what about the numerator?

http://www.wolframalpha.com/input/?i=plot+%283x-1%29%2F%282x-6+%29

ok, so there's only one value of x?

in this case yes..

ok. so how about in this case.
f(x) = 3x-2 if x<0
x^2+x if \[x \ge0\]

http://www.wolframalpha.com/input/?i=piecewise[{{3x-2%2Cx%3C0}%2C{x^2%2Bx%2Cx%3E%3D0}}]