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When x->0, x and sinx are equivalent infinitesimal quantities, so replacing all x's with sinx gives :

sinx->0 lim [arcsin(sinx) / sinx]^

sinx->0 lim (x sinx) gives the logarithm of the limiting formula to get :

sinx->0 lim *ln(x sinx) In the same way, replace all sinx with x:

x->0 lim [1/(1-cosx)]*ln(sinx/x)

x->0 lim [ln(sinx x)] (1-cosx)

x->0 lim [(x/sinx)*(cosx/x - sinx/x^2)] / sinx

x->0 lim (xcosx - sinx) [x(sinx) 2] Ropitta's Rule:

x->0 lim (cosx-xsinx-cosx) / [(sinx)^2+2xsinxcosx]

x->0 lim (-xsinx) / [(sinx)^2+2xsinxcosx]

x->0 lim -1 / [(sinx/x)+2cosx]

Revert logarithm to e (-1 3).

It can be calculated using x=ln (x power of e).

d(δg/t)/dt=(1/t^2)·[tdδg-δgdt)/dt]=-δh/t^2

The d represents the partial differential symbol, which is not easy to play.

It's not just a high number, remember if it's the calculation of entropy in chemistry or something. No, I forgot all about chemistry since I was in graduate school, and you should check more similar textbooks, which should involve approximations and rounding terms. I'm ashamed, I can't help you, I only came in when I saw Gao Shu.

That is, if you put the f(x) represented after ln(1-x) and the taylor of f(x) itself, this is uniqueness.

The uniqueness of McLaughlin is that f(x) McLaughlin series has only one form.

f(n)(0)/n!Only =-1 (n-4).

The indefinite integral of the following equation is corrupted to obtain a0x+(a1)(x*2) 2+...an(x*n+1)/(n+1)+c=0

x=0 gives c=0, x=0, a0x+(a1)(x*2) 2+...an(x*n+1) (n+1)=0, when x=1 is known, a0x+(a1)(x*2) pin withers 2+.an(x*n+1) (n+1)=0, there is a failure of Raul's theorem, and the conclusion is true.

y''=1 y fixed y'=p then y''=dp dx=dp dy * dy dx=dp dy*p p dp dy=1 y pdp=1 y dy 1 2p 2=(1 (1-1 2)) y +c p=soil (4 y+c1) dy dx=soil (4 y+c1) 1 (4 y+c1) dy=soil dx The front integral can be set to u 2=y dy=2udu Then: left=2U (4u+c1)du....1 Set (4u+c1)=v....

2 then: 4U+C1=V 2 U=(V 2-C1) 4 DU=VDV 2 Substituting 1 formula: 2(V 2-C1) 4 V *VDV 2 =(V 2-C1) 4 DV points to obtain:

1 12v 3-c1 4v+c2 Substitute 2 in: 1 12 *(4u+c1) (3 2)-c1 4 * 4u+c1)+c2 and then u= y into the above formula: 1 12 * 4 y+c1) (3 2)-c1 4* (4 y+c1)+c2 So:

The general solution is: (4 y+c1) (3 2)-3c1* (4 y+c1)+c2=soil 12x

Of course not, dy=0 instead of 1

<> is like sending a stupid state to accompany the dust.

<> is stupid to let the game go.