limit as x approaches 7- of (|x-7|)/(x-7)
1) Compute the limit of |x -7|x -7 as x → 7⁻
ⓘ The limit of a sum of finite limits is the sum of the limits
The limit of x -7 is (7 -7)⁻ when x → 7⁻
Let's calculate: 7 -7
ⓘ Start from 7 and move back 7 on the number line, down to zero: 7 -7 = 0
0
ⓘ The function abs is continuous, its limit is its value at the limit of its argument
The limit of |x -7| is 0⁺ when x → 7⁻
For |x -7|x -7, the limit takes the indeterminate form 0/0 when x → 7
2) We need a Taylor expansion for: |x -7|
Apply the Taylor development -X of abs with X=x -7 around A=0 (order N=1)
-(x -7)
3) We redo the calculation using the Taylor expansion: -x + 7
-x + 7x -7
The fraction: -x + 7x -7 simplifies by the common factor: x -7 into the new expression: -1
So the limit of |x -7|x -7 is -1 when x → 7⁻
▶Limit:   -1