my answer is (-m^2+n^2)/2
can anyone verify this?
0/0 so use L'Hopitals
lim (x→0) (cos(mx) - cos(nx))/x²
= lim (x→0) (- msin(mx) + nsin(nx))/2x
= lim (x→0) (nsin(nx) - msin(mx)) /2x
still 0 / 0 so do it again
= lim (x→0) (n²cos(nx) - m²cos(mx))/2
= (n² - m²)/2
so yes you are correct
Case 1 : m = n
=> lim ((x → 0) (cosmx - cosnx)/x^2
= 0
(because cosmx - cosnx = cosmx - cosmx = 0)
Case 2 : m ≠ n
= lim [(x → 0) (2sin(m+n)x/2*sin(n-m)x/2] / x^2
= (n^2 - m^2)/2 lim [(x → 0) [sin(m+n)x/2 * sin(n-m)x/2] / (n^2 - m^2)x^2/4
= (n^2 - m^2) / 2
=> your answer is correct only if m ≠ n.
For m = n, the value of the limit is 0.
If you haven't learned L'Hopital's rule, you can still get it by using trig. identity.
cosmx - cosnx
= cos[(m+n)x/2 + (m-n)x/2] - cos[(m+n)x/2 - (m-n)x/2]
= -2sin[(m+n)x/2]sin[(m-n)x/2]
So,
lim{x->0}(cosmx-cosnx)/x^2
= lim{x->0} -2sin[(m+n)x/2]sin[(m-n)x/2]/x^2
= (-1/2)(m+n)(m-n)
= (-1/2)(m^2 - n^2)
= (1/2)(n^2 -m^2)
By the way, the answer covers the case m = n.
lim x-----> 0 (cos mx - cos nx)/x^2
= lim x------->0 [- 2sin (m+n)x/2 .sin (m-n)x/2]/x^2
= - 2[lim x---->0 sin((m+n)x/2)/x] x [lim x---->0 sin((m-n)x/2)/x]
...............(1)
take, (m+n)x/2 = z => when x----> 0 => z ----> 0
and 2z/(m+n)
similarly, take, (m -n)x/2 = p
so, when x--->0 => p--->0
and x = 2p/(m -n)
now,
lim x----> 0 sin((m+n)x/2)/x
= lim z---->0 sin z/(2z/(m+n))
= lim z---->0 (m+n)/2 . sin z/z
= (m+n)/2 . lim z-->0 sin z/z
=(m+n)/2 [since, lim z---> 0 sin z/z = 1]
similarly,
lim x----> 0 sin((m - n)x/2)/x
= lim p-----> 0 sin p/(2p/(m-n))
= (m - n)/2 lim p ---> 0 sin p/p
= (m - n)/2
so now we have,
lim x --->0 (cos mx - cos nx)/x^2
= - 2[(m+n)/2][(m - n)/2]
= - 2(m^2 - n^2)/4
= - (m^2 - n^2)/2
= (- m^2 + n^2)/2 <==ANSWER
Hope I helped u :)
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Answers & Comments
Verified answer
0/0 so use L'Hopitals
lim (x→0) (cos(mx) - cos(nx))/x²
= lim (x→0) (- msin(mx) + nsin(nx))/2x
= lim (x→0) (nsin(nx) - msin(mx)) /2x
still 0 / 0 so do it again
= lim (x→0) (n²cos(nx) - m²cos(mx))/2
= (n² - m²)/2
so yes you are correct
Case 1 : m = n
=> lim ((x → 0) (cosmx - cosnx)/x^2
= 0
(because cosmx - cosnx = cosmx - cosmx = 0)
Case 2 : m ≠ n
=> lim ((x → 0) (cosmx - cosnx)/x^2
= lim [(x → 0) (2sin(m+n)x/2*sin(n-m)x/2] / x^2
= (n^2 - m^2)/2 lim [(x → 0) [sin(m+n)x/2 * sin(n-m)x/2] / (n^2 - m^2)x^2/4
= (n^2 - m^2) / 2
=> your answer is correct only if m ≠ n.
For m = n, the value of the limit is 0.
If you haven't learned L'Hopital's rule, you can still get it by using trig. identity.
cosmx - cosnx
= cos[(m+n)x/2 + (m-n)x/2] - cos[(m+n)x/2 - (m-n)x/2]
= -2sin[(m+n)x/2]sin[(m-n)x/2]
So,
lim{x->0}(cosmx-cosnx)/x^2
= lim{x->0} -2sin[(m+n)x/2]sin[(m-n)x/2]/x^2
= (-1/2)(m+n)(m-n)
= (-1/2)(m^2 - n^2)
= (1/2)(n^2 -m^2)
By the way, the answer covers the case m = n.
lim x-----> 0 (cos mx - cos nx)/x^2
= lim x------->0 [- 2sin (m+n)x/2 .sin (m-n)x/2]/x^2
= - 2[lim x---->0 sin((m+n)x/2)/x] x [lim x---->0 sin((m-n)x/2)/x]
...............(1)
take, (m+n)x/2 = z => when x----> 0 => z ----> 0
and 2z/(m+n)
similarly, take, (m -n)x/2 = p
so, when x--->0 => p--->0
and x = 2p/(m -n)
now,
lim x----> 0 sin((m+n)x/2)/x
= lim z---->0 sin z/(2z/(m+n))
= lim z---->0 (m+n)/2 . sin z/z
= (m+n)/2 . lim z-->0 sin z/z
=(m+n)/2 [since, lim z---> 0 sin z/z = 1]
similarly,
lim x----> 0 sin((m - n)x/2)/x
= lim p-----> 0 sin p/(2p/(m-n))
= (m - n)/2 lim p ---> 0 sin p/p
= (m - n)/2
so now we have,
lim x --->0 (cos mx - cos nx)/x^2
= - 2[(m+n)/2][(m - n)/2]
= - 2(m^2 - n^2)/4
= - (m^2 - n^2)/2
= (- m^2 + n^2)/2 <==ANSWER
Hope I helped u :)