lim x-->0 [cos(mx) - cos(nx)] / x^2
m and n are constants.
lim [ cos(mx) - cos(nx) ] / (x^2)
x -> 0
This is of the form [ 1 - 1 / 0^2 ], or [0/0], so we may apply L'Hospital's rule. Remember that m and n are constants and should be treated as such.
lim [ -sin(mx)(m) - (-sin(nx))(n) ] / (2x)
Simplifying,
lim [ -m sin(mx) + n sin(nx) ] / (2x)
This is of the from [0 + 0 / 0], or [0/0], so we may apply L'Hospital's rule once again.
lim [ -m (cos(mx))(m) + n cos(nx)(n) ] / (2)
Simplifying again,
lim [ -m^2 cos(mx) + n^2 cos(nx) ] / 2
Now we can safely plug in x = 0 to evaluate the limit.
[ -m^2 cos(0) + n^2 cos(0) ] / 2
[-m^2 + n^2 ] / 2
Final answer is
(1/2)(n^2 - m^2)
= lim [ (-msinmx + nsinnx ) / (2x) ]
= lim [ (-m^2 cosmx +n^2 cosnx ) / 2 ]
= (n^2-m^2)/2
[ some purists might not like the double use of l'hopital but if f''/g'' converges, f'/g' converges to same limit .. without doing a normal proof that should be sufficient for f/g to converge to same limit ]
but you could just have said
cos mx = 1 - (mx)^2/2 + O(x^4)
so cosmx - cos nx = (n^2-m^2) x^2/2 + O(x^4)
from which the same limit is obvious
Comments
lim [ cos(mx) - cos(nx) ] / (x^2)
x -> 0
This is of the form [ 1 - 1 / 0^2 ], or [0/0], so we may apply L'Hospital's rule. Remember that m and n are constants and should be treated as such.
lim [ -sin(mx)(m) - (-sin(nx))(n) ] / (2x)
x -> 0
Simplifying,
lim [ -m sin(mx) + n sin(nx) ] / (2x)
x -> 0
This is of the from [0 + 0 / 0], or [0/0], so we may apply L'Hospital's rule once again.
lim [ -m (cos(mx))(m) + n cos(nx)(n) ] / (2)
x -> 0
Simplifying again,
lim [ -m^2 cos(mx) + n^2 cos(nx) ] / 2
x -> 0
Now we can safely plug in x = 0 to evaluate the limit.
[ -m^2 cos(0) + n^2 cos(0) ] / 2
[-m^2 + n^2 ] / 2
Final answer is
(1/2)(n^2 - m^2)
= lim [ (-msinmx + nsinnx ) / (2x) ]
= lim [ (-m^2 cosmx +n^2 cosnx ) / 2 ]
= (n^2-m^2)/2
[ some purists might not like the double use of l'hopital but if f''/g'' converges, f'/g' converges to same limit .. without doing a normal proof that should be sufficient for f/g to converge to same limit ]
but you could just have said
cos mx = 1 - (mx)^2/2 + O(x^4)
so cosmx - cos nx = (n^2-m^2) x^2/2 + O(x^4)
from which the same limit is obvious