color(blue)("Determine vertical asymptotes")
Mathematically you are not allowed to divide by 0. This situation is called 'undefined'.
So at x=2 the equation is undefined. Thus x=2 is an asymptote.
Suppose x>2 then 2-x < 0
As x becomes increasingly closer to 2 then 2-x becomes smaller and smaller but still negative.
In the same way, when x<2 but approaching 2 then 2-x becomes smaller and smaller but is still positive.
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So 1/(2-x) becomes increasingly larger but negative or positive depending on what side of 2 we find x to be.
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When writing limits; when approaching a value from the positive side then for the value 2 it would be written as color(white)()^(+)2
Same way for approaching from the negative side it is color(white)()^(-)2
Thus
y = lim_(xtocolor(white)()^(+)2)1/(2-x) = - oo " "larr ( x > 2)
y = lim_(xtocolor(white)()^(-)2)1/(2-x) = + oo " "larr (x<2)
color(green)("So "x=2 " is the vertical asymptote")
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color(blue)( "Determine horizontal asymptotes")
y = lim_(xtocolor(white)()^(+) oo)1/(2-x) = 1/(2-oo) =- 1/oo = color(white)()^(-) 0
y = lim_(xtocolor(white)()^(-) oo)1/(2-x) = 1/(2+oo) = +1/oo = color(white)()^(+) 0
color(green)("So "y=0 " is the horizontal asymptote")
Tony B