How do you find vertical, horizontal and oblique asymptotes for #y=1/(2-x)#?

1 Answer
May 31, 2016

Answer:

Vertical asymptote #x=2#
Horizontal asymptote #y=0#

Explanation:

#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