asymptotes of 1/(x-1)^2
Let's find the asymptotes of 1(x -1)2, for real x
Vertical asymptotes: find the excluded values, where the denominator is zero. Then check whether the curve goes to infinity near these values
Let's solve in ℝ: (x -1)2 = 0
(x -1)2 = 0 is equivalent to: x -1 = 0
ⓘ We add 1 to both sides
Solution found: x = 1
As x approaches 1 from the left (smaller values), the function tends to +∞; from the right (larger values), it tends to +∞
The function tends to infinity on at least one side: the line x = 1 is therefore a vertical asymptote
At infinity, the numerator degree is smaller than the denominator degree: the line to test is y = 0
For a horizontal asymptote, the function must tend to a finite value: check the limit of 1(x -1)2 at both infinities
As x tends to +∞, 1(x -1)2 tends to 0; as x tends to -∞, 1(x -1)2 tends to 0
The horizontal asymptote is y = 0; there is no slant asymptote
▶Asymptotes:   { x = 1, y = 0 }
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