Finding limit x → Alim (LNX LNA) / x-a Seeking the limit x→a lim (lnx-lna)/(x-a)

Finding limit x → Alim (LNX LNA) / x-a Seeking the limit x→a lim (lnx-lna)/(x-a)

It's a 0 / 0 uncertainty, using the L'Hospital Rule
Original limit = Lim {d [LNX LNA] / D [x-a]} = Lim {(1 / x) / 1} = 1 / A

When x → a, Lim f (x) = + ∞, when x → + ∞, Lim g (x) = a, it is proved that when x → a, Lim g (f (x)) = a

Let f (x) = t
That is to say, when x → a, Lim g (T) = a
From the problem we know that when x → + ∞, Lim g (x) = a
So t = + ∞ so f (x) = + ∞
Because when x → a, Lim f (x) = + ∞
So the original proposition is proved

If Lim [(α (x) / β (x)] x → a exists and lim α (x) = 0, X → a, it is proved that LIM β (x) = 0

If Lim [(α (x) / β (x)] ≠ 0, then we can let α (x) / β (x) = a (x), and then let Lim α (x) / β (x) = Lim a (x) = a have α (x) / A (x) = β (x), take Lim β (x) = 0 / a = 0 on both sides. If a is 0, then there is a counterexample α (x) = 0, β (x) = 1 / x, Lim α (x) / β (x) = Lim when x → 0

lim(x→0,y→0） xy/（√2-e^xy）-1=?
Such as the title

It should be: LIM (x - > 0, Y - > 0) XY / [√ (2-e ^ XY) - 1] this is a 0 / 0 type limit formula, using the lobita rule formula of binary function limit: LIM (x - > x0, Y - > Y0) [f (x, y) / g (x, y)] = LIM (x - > x0, Y - > Y0) {[f'x (x, y) DX + f'y (x, y) dy] / [g'x (x, y) DX + g'y (x, y) dy]} where DX = x-x0