Limit x → 0 [(1 + MX) ^ n - (1 + NX) ^ m] / x ^ 2 (n, M is a positive integer)

Limit x → 0 [(1 + MX) ^ n - (1 + NX) ^ m] / x ^ 2 (n, M is a positive integer)

0 / 0, lobida twice, and the molecular denominator is derived separately
1:[n*m*(1+mx)^(n-1)-m*n*(1+nx)^(m-1)]/2x
2:[n*m^2*(n-1)(1+mx)^(n-2)-m*(m-1)*n^2*(1+nx)^(m-2)]/2
=m^2*n(n-1)-n^2*m(m-1)=mn(n-m)

Find the limit [(1 + NX) ^ (1 / M) + (1 + MX) ^ (1 / N)] / x, X - > 0, where m, n are positive integers, and the infinite limit of X is close to 0. I used two derivatives to get N.m ^ 2 (n-1) - (m-1). M.n ^ 2 / 2. How did Mn. (n-m) come from

If you don't use series expansion, there are other ways. Have you studied the derivative of power function
y=(1+nx)^(1/m)-(1+mx)^(1/n)
Find the derivative of Y at x = 0, and the result is the answer to your question
The following is to use the series expansion method to find the limit:
If it's a plus sign, the limit is infinite. If it's a minus sign, it's a limit of type 0 / 0. If you don't use the lobita rule, it's very easy to use the series expansion method
The series (1 + x) ^ n can be expanded as: 1 + NX +
lim [(1+nx)^(1/m)-(1+mx)^(1/n)]/x
=lim [(1+nx/m)-(1+mx/n)]/x
=lim [(nx/m)-（mx/n)]/x
=(n/m)-（m/n)=(n-m)(n+m)/(nm)

What are cos (10 * pi / 180) and sin (10 * pi / 180) / / respectively?

cos(10*PI/180)
= cos(PI/18)
= 0.9848
sin(10*PI/180)
= sin(PI/18)
= 0.1736