Is LIM (e ^ x) x zero when it tends to negative infinity?

Is LIM (e ^ x) x zero when it tends to negative infinity?

When x approaches negative infinity, e ^ - X and X both approach infinity, and the law of lobita can be used
lim(e^x)x=lim(x)/(e^-x)=lim1/(-e^-x)=0

X tends to 0, Lim f (x) / x = 1, f '' (x) > 0. It is proved that f (x) > X

∵f(x)=x+o(x) ∴f'(0)=1 f(0)=0
Write out the McLaughlin formula expansion of F (x)
F (x) = f (0) + F '(0) x + F' '(m) x ^ 2 m is between 0 and X
f(x)=x+f''(m)x^2
∵f''(x)>0
f(x)-x>0

F (x) is continuous at x = 2, Lim [f (x) / (X-2)] = 3 (x tends to 2), find f (2) and f '(2)
F (x) is continuous at x = 2, Lim [f (x) / (X-2)] = 3 (x tends to 2), find f (2) and f '(2)

3 = Lim [f (x) / (X-2)] (x tends to 2) = Lim [f '(x)] (x tends to 2) = f' (2) 0 / 0 type limit 3 = Lim [f (x) / (X-2)] (x tends to 2) can get 1 = limf (x) / [3 (X-2)] (x tends to 2) so f (2) = Lim [f (x)] (x tends to 2) = Lim [3 (X-2)] (x tends to 2) = 0