What is the limit of xlnx + X when x tends to zero

What is the limit of xlnx + X when x tends to zero

Xlnx + x = (1 + LNX) / [1 / x], X → 0, is ∞ / ∞ type, can use the law of lobita
limx→0(1/x)/[-1/x^2]=-limx→0(x)=0

The limit X of xlnx tends to 0

The answer is zero
The original formula is equal to LNX divided by 1 / x, and the denominator is infinite. By using L, hospital's rule, the derivative result is - x, X tends to 0, then - x = 0, and the limit is 0

The limit of xlnx tending to zero
The limit of xlnx tending to zero
The limit of e ^ X / x ^ 3 x tending to positive infinity
sorry
Y = e ^ xsin (x / 2), find the derivative of Y
Y = (e ^ x + e ^ - x) ^ 100, find the derivative of Y

The limit of xlnx tending to 0 = 0 (using the law of Robita)
The limit of e ^ X / x ^ 3 x tends to be positive infinity (continuous three times according to the law of Robita) dy / DX = e ^ xsin (x / 2) + (e ^ xcos (x / 2)) / 2,
dy/dx=(100*(e^x + e^-x)^100)*(e^x - e^-x)