It is known that the image of function f (x) is symmetrical about the origin and passes through the origin. When x > 0, f (x) = x (x-1), the expression of F (x) is obtained

It is known that the image of function f (x) is symmetrical about the origin and passes through the origin. When x > 0, f (x) = x (x-1), the expression of F (x) is obtained

F (x) = absolute value of x ^ 2-x

x→∞ lim(3x+sinx)/(2x-sinx)=

Original formula = LIM (x - > ∞) (3 + SiNx / x) / (2-sinx / x)
Because SiNx is bounded, 1 / X is infinitesimal
So when X - > ∞, SiNx / X - > 0
So the original formula = 3 / 2

Find LIM (n → 0) (e ^ x-e ^-x-2x) / (x-sinx)

=2
Robida's law, just take the derivative of the numerator and denominator

lim[x→∞]【(x+1)sinx】/(2x^3-3x+2)

|sinx|∞)(x+1)/(2x^3-3x+2) =0
=>lim(x->∞)【(x+1)sinx】/(2x^3-3x+2) =0

Calculate the limit of the following function: (1) Lim x → e (xlnx + 2x) (2) Lim x → π / 2 (SiNx / 2cos2x)

(1)
lim(xlnx+2x)=elne+2e=e+2e=3e
(2)
lim[sinx/(2cos2x)]=sin(π/2)/[2cos(2*π/2)]=1/[2*(-1)]=-1/2

LIM (x → 0) (2 + x) SiNx ~ 2x why are these two limits equivalent?

lim(2+x)sinx/(2x)
=lim[(2+x)/2]*sinx/x
=(2+0)/2*1
=1
So equivalent