L'Hospital x-->0; Could you correct my attempt?

Question

x-->0 (cos(mx) - cos(nx)) /x^2
because 0/0: L'Hosp.

x-->0 -m*sin(mx)m + n*sin(nx) /2x
because 0/0: L'Hosp.

x-->0 [(n^2)(cos(nx))+sin(nx)-(m^2)cos(mx)-sin(mx)]/2
reduce:
x-->0 [(n^2) -(m^2)(-1)]/2

[(n^2) + (m^2)]/2 <---***Wrong!***

doh! lolololol
Thank you :0)

actually, on closer inspection I could npt find the (-),,,, Thanks Dr D!

Answer 1

Initially
[cos(mx) - cos(nx)] / x^2

LH: [-m*sin(mx) + n*sin(nx)] / (2x)

LH: [-m^2 *cos(mx) + n^2*cos(nx) ] / 2

x = 0
limit = (n^2 - m^2) / 2

Answer 2

x-->0 (cos(mx) - cos(nx)) /x^2
because 0/0: L'Hosp.

x-->0 -m*sin(mx) + n*sin(nx) /2x
because 0/0: L'Hosp.

x-->0 [-(m^2)(cos(mx))+n^2cosin(nx)/2
= [(n^2) -(m^2)]/2 <-- Right

[(n^2) + (m^2)]/2 <---***Wrong!***

Answer 3

ANSWER: [(n²) - (m²)]/2

2nd step: remove the m after sin(mx)

3rd step: x-->0 { -m² cos(mx) + n² cos(nx) }/2
... just use another L'Hôpital, right?


d: