6x^2cotx-conec2x

"lim x→0" [x∙csc(2x)] / cos(5x)

>>> trig: csc(a) = 1/sin(a)

= "lim x→0" x / [sin(2x) ∙ cos(5x)]
= "lim x→0" x / [2sin(x)cos(x) ∙ cos(5x)]

>>> note the 1/2 inside the limit

= "lim x→0" {1/2} ∙ {x / [sin(2x) ∙ cos(5x)]}

>>> take {1/2} out of the limit

= {1/2} "lim x→0" {x / [sin(2x) ∙ cos(5x)]}

now, the limit becomes zero, even though, technically the entire limit is zero at x=0, but since that is the ONLY point where the limit is zero, the gap is bridged with the {1/2} taken out of it

= 1/2         {answer}



"lim x→0" [x∙csc(2x)] / cos(5x) = 1/2