According to the definition, it is proved that when x tends to 0, the function y = (1 + 2x) / X is infinite. Q. what conditions can x satisfy to make the absolute value of Y greater than 10000

Y = (1 + 2x) / X is reduced to y = 2 + 1 / X. now y is greater than 10000, which is 10000

It is proved that the square + 2x of the function y = x is an increasing function on [0, positive infinity]

Let X1 > x2 > = 0
Then y1-y2 = (x1) ^ 2 + 2x1 - (x2) ^ 2-2x2
=(x1+x2)(x1-x2)+2(x1-x2)
=(x1+x2+2)(x1-x2)>0
That is, Y1 > Y2
So the square of y = x + 2x is an increasing function on [0, positive infinity]

LIM (x tends to 0 +) x ^ TaNx find the limit?

LIM (x tends to 0 +) x ^ TaNx
=E ^ LIM (x tends to 0 +) LNX ^ TaNx
=E ^ LIM (x tends to 0 +) LNX * TaNx
=E ^ LIM (x tends to 0 +) LNX / Cotx (∞ / ∞)
=E ^ LIM (x tends to 0 +) (1 / x) / (- CSC ^ 2x)
=E ^ LIM (x tends to 0 +) - SiNx
=e^0
=1

TaNx minus x is the limit of the power of X multiplied by 2 when x tends to 0 It seems to be the best choice for advanced mathematics

F (x) = (tanx-x) / (2x ^ 3) first observe that x → 0 is the fraction, both up and down tend to 0. Use the robbida's law to calculate the derivatives of the fraction, and the final result is 1 / 6. That's about it

X tends to 0, find the limit of 1 / x ^ 2-1 / (x * TaNx)

Equal to 1

Find the limit LIM (e ^ x) - (e ^ - x) - 2x / (tanx-x) x tends to 0

The formula of the last division is lobida's law = Lim e Λ X - lime Λ - x + Lim 2 / (SEC Λ 2 x - 1) = 1 + 1 + 0 = 2

According to the definition, it is proved that when x tends to 0, the function y = (1 + 2x) / X is infinite. Q. what conditions can x satisfy to make the absolute value of Y greater than 10000