Lim x →∞ xsin (1 / x)

Lim x →∞ xsin (1 / x)

The limit LIM (xsin * 2 / x + 2 / X * SiNx) when x tends to zero

I'm wrong
In fact, it is known that when x → 0, Lim [xsin (1 / x)] = 0
lim(xsin*2/x+2/x*sinx)
=lim(xsin*2/x)+lim(2/x*sinx)
=2lim[(x/2)sin*2/x]+2lim(sinx/x)
=0+2
=2

Find LIM (x ^ 3 (sin1 / X-1 / 2 * sin2 / x)) x tends to infinity

LIM (x ^ 3 (sin1 / X-1 / 2 * sin2 / x)) x tends to infinity
=LIM (sint-1 / 2 * sin2t) / T ^ 3T tends to 0
=LIM (cost-cos2t) / 3T ^ 2T tends to zero
=LIM (- Sint + 2sin2t) / 6T tends to 0
=LIM (- cost + 4cos2t) / 6T tends to 0
=(-1+4)/6
=1/2