limit m tends to n, m square sin m minus n square sin n divided by m minus n

Question

lim m tends to n, m square sin m minus n square sin n divided by m minus n

Answer 1

Lim m→n of (m²sin(m)-n²sin(n))/(m-n).

Let m=n+h where h is very small.

The expression becomes:

[(n+h)²sin(n+h)-n²sin(n)]/h,

[(n²+2nh+h²)(sin(n)cos(h)+cos(n)sin(h))-n²sin(n)]/h. 

h² can be ignored (as can 2nh²cos(n)) because h is small, and sin(h)=h (approx) and cos(h)=1 (approx).

So we have:

[(n²+2nh)(sin(n)+hcos(n))-n²sin(n)]/h=

[n²sin(n)+n²hcos(n)+2nhsin(n)-n²sin(n)]/h=

n²cos(n)+2nsin(n) for general n as m→n.

Note that this expression is the derivative wrt n of n²sin(n).