How to find LIM (x tends to 0 +) LNX / 2x?

How to find LIM (x tends to 0 +) LNX / 2x?

lim(x→0+)lnx/（2x）=-∞,
Because LNX → - ∞, - ∞ / 0 + → - ∞

LIM (x --- infinity) (3x's square + 2x + 4) parts (2x's Square - 3x + 5)

Divide by X & sup2;
=(2-3/x+5/x²)/(3+2/x+4/x²)
X tends to infinity, so x tends to zero at the end of the denominator
So the limit = (2-0 + 0) / (3 + 0 + 0) = 2 / 3

Limit problem Limx → 0 sin3x / sin5x =?

When x → 0, sin3x → 0, similarly, sin5x → 0; when x → 0, the numerator denominator tends to 0 at the same time. So we can get the answer 3 / 5 by taking the derivative of the numerator denominator and substituting x = 0. Or we can get the answer 3 / 5 directly by using the equivalent infinitesimal, when x → 0, sin3x ~ 3x; sin5x ~ 5x