LIM (e ^ x-1) / X when x tends to 0 If we use the law of Roberta to solve the problem, is it to bring (e ^ x-1) / x into the (u'v-uv ') / V ^ 2 formula?

LIM (e ^ x-1) / X when x tends to 0 If we use the law of Roberta to solve the problem, is it to bring (e ^ x-1) / x into the (u'v-uv ') / V ^ 2 formula?

LIM (e ^ x-1) / X when x tends to zero
=lim(x->0)(e^x-0)/1
=lim(x->0)(e^x)
=e^0
=1
It's not your formula, it's the derivation of the numerator and denominator

If f (x) is differentiable at x = a, find Lim X - > a XF (a) - AF (x) / x-a

[xf(a)-af(x)]/(x-a)=[(x-a)f(a)+a(f(a)-f(x))]/(x-a)=f(a)+a[f(a)-f(x)]/(x-a),
==> lim x->a {[xf(a)-af(x)]/(x-a)}
=f(a)+lim x->a {a[f(a)-f(x)]/(x-a)}
=f(a)-a lim x->a {[f(x)-f(a)]/(x-a)}
=f(a)-af'(a).

If Lim [f (x) / g (x)] = a, is Lim [g (x) / F (x)] equal to 1 / a? Just like infinity and infinitesimal, we can invert the numerator and denominator. If DX / dy = 1 / y ', can we consider dy / DX = y'?
The most important thing is: why? Please, great Xia, what do you think and analyze? Why? Or why not?
The concept is really confusing, please give me more advice ~ thank you very much~
If DX / dy = 1 / y 'is equivalent to dy / DX = y', how can we use dy / DX = y 'to deduce the conclusion d ^ 2x / dy ^ 2 = - y' '/ (y') ^ 3?

Yes, the conclusion is correct ∵ Lim [g (x) / F (x)] = Lim [f (x) / g (x)] - & sup1; = {Lim [f (x) / g (x)]} - & sup1; = 1 / A. for example, y = LNX, x = E & # 710; y, DX / dy = E & # 710; y, dy / DX = 1 / x, E & # 710; y = E & # 710; LNX = x, don't you come out
D ^ 2x / dy ^ 2 = D (1 / y ') / dy = [D (1 / y') / DX] / (DX / dy) insert a DX in the middle, because y 'is a function of X, the upper side of the fraction is derived by division, which is equal to [1' * y '- 1 * (y') '] / (y') & # 710; 2 = - y '' / (y ') ^ 2, the lower side of the fraction is equal to DX / dy = y', and then the numerator is divided by the denominator = - y '' / (y ') ^ 3
There should be a condition in the second conclusion that the derivative of the inverse function is not equal to zero