lim(x->∞) (sin2x)/(x^2)= lim (sin2x)/(x^2)=_______ (x->∞)

lim(x->∞) (sin2x)/(x^2)= lim (sin2x)/(x^2)=_______ (x->∞)

lim(sin2x)/(x^2)=lim[(sin2x)*1/(x^2)]
When X - > ∞, sin2x is a bounded function and 1 / (x ^ 2) is infinitesimal
The product of bounded function and infinitesimal is infinitesimal
So the result is infinitesimal, which is zero

lim(x->0)(x^2+2x)cos5x/sin2x

Obviously, when x tends to 0, cos5x tends to 1
that
Original limit
=lim(x->0) (x^2+2x) / sin2x
Nobida's law
=lim(x->0) (2x+2) /2cos2x
When x tends to 0, cos2x tends to 1
so
Original limit
= 1

lim(x→0)（x-sin2x）/（x+sin5x）
How to find this limit? Please give me the detailed steps, I'm a beginner, master ask

X → 0, sin2x → 2x, sin5x → 5x, and then it's easy to do