How to do LIM (x → 0) (TaNx SiNx) / (X & # 179;)

How to do LIM (x → 0) (TaNx SiNx) / (X & # 179;)

It's better to simplify the law of lobita
Original formula = LIM (SiNx / cosx -- SiNx) / x ^ 3
=lim sinx/x* lim (1--cosx)/(x^2*cox)
=1*lim (1--cosx)/x^2
=1*lim sinx/(2x)
=1*1/2
=1/2.

lim x->0 [(1+tanx)^1/2-(1+sinx)^1/2]/[xln(1+x)-x^2]
When x tends to zero, the difference between the square root of the sum of the tangent of one plus X and the square root of the sum of the sine of one plus x divided by the quotient of X multiplied by the product of the logarithm of the sum of one plus x with E as the base minus the difference of the square of X, the limit is?

If you multiply (1 + TaNx) ^ 1 / 2 - (1 + SiNx) ^ 1 / 2,
The molecule becomes (1 + TaNx) - (1 + SiNx) = TaNx SiNx
The denominator becomes [XLN (1 + x) - x ^ 2] * [(1 + TaNx) ^ 1 / 2 + (1 + SiNx) ^ 1 / 2]
Original formula = Lim [(TaNx SiNx) / {[XLN (1 + x) - x ^ 2] * [(1 + TaNx) ^ 1 / 2 + (1 + SiNx) ^ 1 / 2]}
】
=LIM【(tanx-sinx)/[x(ln(1+x)-x)]】*LIM【1/[（1+tanx)^1/2+(1+sinx)^1/2]}】
(from the continuity of the function, the limit of the latter term is 1 / 2)
Ask below
LIM【(tanx-sinx)/[x(ln(1+x)-x)]】
because
tanx-sinx=sinx/cosx-sinx*cosx/cosx=sinx(1-cosx)/cosx
therefore
LIM【(tanx-sinx)/[x(ln(1+x)-x)]】=LIM【sinx(1-cosx)/[x(ln(1+x)-x)]/cosx】
=Lim [x ^ 3 / (2x (LN (1 + x) - x)) * 1 / cosx] (based on equivalent infinitesimal)
=Lim [x ^ 2 / (LN (1 + x) - x)] * LIM (1 / cosx) (obviously the latter limit is 1)
=Lim [2x / [1 / (1 + x) - 1]] (lobida's law)
=-1
So the original formula = - 1 / 2

If one sixth of a is greater than one seventh of B, then a must be greater than B

∵a/6>b/7,
A > 6 / 7 * B is not necessarily greater than B
For example, when a = 7 and B = 7, a / 6 > b / 7 holds, but a = 7 > b = 7 does not