Limx tends to positive infinity [1 / (1-x) - 1 / (1-x ^ 3)] to find function limit

Limx tends to positive infinity [1 / (1-x) - 1 / (1-x ^ 3)] to find function limit

The original formula = LIM (x →∞) (1 + X + X & # 178;) / (1-x) (1 + X + X & # 178;) - 1 / / (1-x) (1 + X + X & # 178;)] = LIM (x →∞) (x + X & # 178;) / (1-x & # 179;) up and down divided by X & # 179; = LIM (x →∞) (1 / X & # 178; + 1 / x) / (1 / X & # 179; - 1) = (0 + 0) / (0-1) = 0

Finding the limit limx tends to 2 1 / (X-2) - 12 / (x ^ 3-8)

1 / (X-2) - 12 / (x ^ 3-8) = (x ^ 2 + 2x + 4-12) / [(X-2) (x ^ 2 + 2x + 4)] = (x ^ 2 + 2x-8) / [(X-2) (x ^ 2 + 2x + 4)] = (X-2) (x + 4) / [(X-2) (x ^ 2 + 2x + 4)] = (x + 4) / [(x ^ 2 + 2x + 4)] = (x + 4) / (x ^ 2 + 2x + 4) when x tends to 2, the value limit of (x + 4) / (x ^ 2 + 2x + 4) when x = 2 = 1 / 2

The solution of function limit limx tends to 3 (x-1) / (x-3)=

The molecule is a constant, the denominator is 0, and the limit is ∞

Limx tends to the limit of 1, X / 1-x

lim(x→1)[x/(1-x)] = ∞

Let a be a constant, when x tends to infinity, the limit of x [(1 + (1 / x)) ^ A-1]

The limit of X / (x-a) when x tends to a?

There is no limit
Because (x-a) / X has a limit of 0 when x tends to a
So when x tends to a, X / (x-a) tends to infinity
So the original form
When x tends to a, X / (x-a) has no limit

What is the limit of X / SiNx when x tends to infinity

X tends to infinity
SiNx oscillates in [- 1,1]
And molecules tend to infinity
So x / SiNx goes to infinity
So there is no limit

Find the limit! (a ^ x-1) / x, the limit when x tends to 0, it's better to have a process,

When a = 0, the numerator is constant = - 1, and the denominator x tends to 0,
So there is no limit
When a is not equal to 0
According to the law of lobita:
Limit (a ^ x-1) / x = limit a ^ xlna / 1 = LNA
So the limit is LNA

(1 + 2x) ^ X / 1, X tends to the limit of 0 detailed steps

The following figure provides two solutions: Click to enlarge, the screen will be enlarged and then enlarged

When x tends to + ∞, find the limit of (x + e ^ x) ^ (1 / x),

Let: y = (x + e ^ x) ^ (1 / x) LNY = [ln (x + e ^ x)] / Xlim (x →∞) LNY = LIM (x →∞) (1 + e ^ x) / / (x + e ^ x) / /: be an infinitive of type ∞ / ∞, and use the lobita rule; = LIM (x →∞) e ^ X / (1 + e ^ x) / / /: be an infinitive of type ∞ / ∞, and then use the lobita rule; = LIM (x →∞) e ^ X / e ^ x = 1 to get: l