lim x-->3- (x^2-3x) /(x^2-6x+9)

Apply L'Hopital's rule

differentiate the numerator and denominator and substitute x=3

= lim x-->3- (2x-3) /(2x-6)

Numerator approaches 3 while denomanator approaches 0 but negative

The limit is -infinity

lim x-->2pi+ x cot x

= lim x-->2pi+ cot x

= lim x-->2pi+ x * lim x-->2pi+ 1/tan(x)

= 2pi *(infinity)

= infinity

Lim (x² - 3x) / (x² - 6x + 9)

x → 3-

Lim x.(x - 3) / (x - 3)²

x → 3-

Lim x / (x - 3) → when x approaches 3-, the value (x - 3) approaches 0-

x → 3-

Lim x / (x - 3) = 3 / 0- = - ∞

x → 3-

Lim x.cot(x) → when x approaches 2π+, it’s similar when x approaches 0+

x → 2π+

Lim x.cot(x) → when x approaches 0+, then tan(x) approaches 0+, so cot(x) approaches + ∞

x → 2π+

Lim x.cot(x) = + ∞

x → 2π+

lim x→3-, (x^2 - 3x)/(x^2 - 6x + 9) =

lim x→3-, x(x - 3) / (x - 3)^2 = x/(x - 3) = 3/0- = -∞

lim x→2𝜋+, x cotx =

lim x→2𝜋+, x cosx/sinx = (2𝜋+) * 1 / 0+ = +∞