Sin MX divided by sin NX to find the limit of X tending to 0

Sin MX divided by sin NX to find the limit of X tending to 0

When x tends to zero, sin MX is equal to MX, so the answer is m / n

X = 0 is the removable discontinuity of function f (x) = xsin (1 / x). Why? It is clear that both the left and right limits are equal to 0 and equal to f (0) = 0!

First of all, f (0) is undefined, because x is the denominator and cannot be 0, so x = 0 is the discontinuity point. In addition, the limit at 0 exists and is equal, so the discontinuity point can be removed!
Let XO be the discontinuous point of function f (x), then if f (x -) and f (x +) exist, then XO is called the first kind of discontinuity of F (x). If f (x -) = f (x +) and not equal to f (XO) (or F (XO) is undefined), then XO is called the removable discontinuity of F (x)

Let f (x) = sin (NX / 6 + π / 5), where n ≠ 0
1. What is the value of X, the maximum and minimum of F (x) are obtained, and the minimum positive period T is obtained
2. Try to find the minimum positive integer n, so that when the independent variable x changes between any two integers (including the integer itself), the function f (x) has at least one maximum and minimum value

Because when (NX / 6 + π / 5) = π / 2 + 2 π K, f (x) = max = 1
so x=12kπ/n+3π/n-6π/5n
x=(60kπ+9π)/5n f(x)=max
Because when (NX / 6 + π / 5) = - π / 2 + 2 π K, f (x) = min = - 1
so x=12kπ/n-3π/n-6π/5n
f(x)=min
T=2π/(n/6)=12π/n
Because there must be at least one maximum and minimum between two numbers
To make t