LIM (x tends to positive infinity) [x] * sin (1 / x) =?

LIM (x tends to positive infinity) [x] * sin (1 / x) =?

LIM (1 + SiNx sin (SiNx)) ^ x ^ - 3 x tends to 0

lim(x->0) (sinx-sin(sinx))/x^3 =lim(x->0) (cosx-cos(sinx)cosx)/3x^2=lim(x->0) (-sinx+sin(sinx)cos^2x+cos(sinx)sinx)/6x= 1/6*lim(x->0) {[(-sinx)/x] + [sin(sinx)cos^2 x]/x+[cos(sinx)sinx)/x]}= 1/6* { -1...

LIM (n tends to infinity) (x-a) sin (1 / A-X)

When x tends to infinity, then 1 / (A-X) tends to 0
So the original formula = Lim [- sin (1 / A-X) / (1 / A-X)]
=-This is obtained by using one of the two important limits in Higher Mathematics: LIM (x tends to 0) (SiNx / x) = 1. The proof of this limit is available in higher mathematics books. It can be proved by using definitions, inequalities, and the law of Robida