limit as x approaches 0+ of x/(x+|x|)
1) Compute the limit of xx + |x| as x → 0⁺
ⓘ The function abs is continuous, its limit is its value at the limit of its argument
The limit of |x| is 0⁺ when x → 0⁺
ⓘ The limit of a sum of finite limits is the sum of the limits
The limit of x + |x| is 0⁺ when x → 0⁺
For xx + |x|, we encounter the indeterminate form 0/0 as x tends to 0
2) We look for a Taylor expansion of the expression: |x|
Apply the Taylor development X of abs with X=x around A=0 (order N=1)
3) Let's restart using the Taylor expansion: x
xx + x
ⓘ Like terms are grouped by adding their coefficients: x + x = 2x
x2x
ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor x: x2x = 12
12
So the limit of xx + |x| is 12 when x → 0⁺
▶Limit:   12
4) The numeric value of: 12 equals:
▷ 0.5