Find limx [ln (x + 1) - ln (x-1)]; when x approaches infinity

Find limx [ln (x + 1) - ln (x-1)]; when x approaches infinity

Ln (x + 1) - ln (x-1) = ln (x + 1) / (x-1) = ln [1 + 2 / (x-1)] when x tends to infinity, 2 / (x-1) tends to 0, so ln (x + 1) / (x-1) = ln [1 + 2 / (x-1)] tends to ln1 = 0

Limx - > 0 ln x times ln (1 + x)

Is LIM (x - > 0 +) ln x * ln (x + 1)
= = = = = = = = =
Let t = ln (x + 1),
Then x = e ^ t - 1,
And when X - > 0 +,
t ->0+ .
So the original formula = LIM (T - > 0 +) t * ln (e ^ t - 1)
=lim (t ->0+) ln (e^t -1) / (1/t).
According to the law of lobita,
Original formula = LIM (T - > 0 +) [e ^ t / (e ^ t - 1)] / (- 1 / T ^ 2)
= -lim (t ->0+) (t^2 *e^t) /(e^t -1)
= -lim (t ->0+) (2t* e^t +t^2 *e^t) /(e^t)
= -lim (t ->0+) (2t +t^2)
= 0.
= = = = = = = = =
Solution 2: because when X - > 0 +,
ln (1+x) x,
So the original formula is LIM (x - > 0 +) x ln X
=lim (x ->0+) ln x /(1/x).
According to the law of lobita,
Original formula = LIM (x - > 0 +) [(1 / x) / (- 1 / x ^ 2)]
= -lim (x ->0+) x
= 0.

Find the limit of (1-x) Tan (Wu / 2) x when X - > 1

Solution 1: the original formula = LIM (x - > 1) [(1-x) sin (π X / 2) / cos (π X / 2)] = {LIM (x - > 1) [sin (π X / 2)]} * {LIM (x - > 1) [(1-x) / cos (π X / 2)]} = 1 * {LIM (x - > 1) [(1-x) / cos (π X / 2)]} = LIM (x - > 1) [(1-x) / sin (π / 2 - π X / 2)] (using the induced formula) = LIM (x - > 1) [(1-x)]

Limx → + ∞ e ^ x-e ^ - X / E6X + e ^ - X: 2, limx → 0x arcsinx / x ^ 3: 3, limx → 1 (2-x) ^ Tan π X / 2;

1. Multiply up and down by e ^ - x
two
LIM (x → 0) (x-arcsinx) / x ^ 3 (0 / 0, lobita rule)
=LIM (x → 0) [1-1 / √ (1 + x ^ 2)] / (3x ^ 2) (general)
=LIM (x → 0) [√ (1 + x ^ 2) - 1] / [√ (1 + x ^ 2) * (3x ^ 2)] (limit algorithm)
=lim(x→0) [√(1+x^2)-1]/(3x^2) *lim(x→0)1/√(1+x^2)
=LIM (x → 0) [√ (1 + x ^ 2) - 1] / (3x ^ 2) (molecular equivalent infinitesimal)
=lim(x→0) 1/2x^2/(3x^2)
=1/6
three
This is of type 1 ^ oo. You can use the important limit criterion to solve the problem. The details are as follows:
LIM (2-x) ^ Tan (π x) / 2 when x → 1 = Lim [1 + (1-x)] ^ 1 / (1-x) * (1-x) * Tan (π x) / 2 when x → 1 = e ^ LIM (1-x) * Tan (π x) / 2 when x → 1
When the limit x → 1, LIM (1-x) * Tan (π x) / 2 is 0 * OO type, which can be transformed into 0 / 0 or OO / OO type to use Robida
When x → 1, LIM (1-x) * Tan (π x) / 2 = x → 1, Lim [Tan (π x) / 2] / [1 / (1-x)]
=When x → 1, Lim π / 2 * (1-x) ^ 2 / [cos (π x) / 2] ^ 2 = x → 1, lim - π / 2 * 2 (1-x) / [- 2Sin (π x) / 2 * cos (π x) / 2] = x → 1, lim - 2 * (x-1) / [2Sin (π x) / 2 * cos (π x) / 2] = x → 1, lim - 2 * (x-1) / sin (π x) = x → 1, lim - 2 / π cos (π x) = 2 / π, so the limit of the original formula is e ^ 2 / π

The limit of (x - π) Tan (x / 2) when limx → π

lim (x-π) tan(x/2)
= lim (x-π) cot(π/2 - x/2 )
= lim (x-π) / tan (π-x)/2
= lim -2*[ (π-x)/2 ] / tan (π-x)/2
= -2

Find the limit limx → 0 = (1 / x) ^ (Tan x)

Logarithm
Ln primitive = LIM (x → 0) tanxln (1 / x)
=-lim(x→0)lnx/cotx
=-LIM (x → 0) (1 / x) / (- 1 / sin ^ 2 (x)) (Law of lobita)
=lim(x→0)sin^2(x)/x
=0 (sinx~x)
So the original formula = e ^ 0 = 1

Find the limit limx to be close to 0 (1-cosx) / (1-e ^ x)

X → 0, Cox → 1, e ^ x → 1, so the denominator tends to 0
So you can use the lobita rule
Derivation of numerator and denominator respectively
Original limit = limx → 0 (SiNx / - e ^ x) = 0 / - 1 = 0

Let limx go to 0, find the limit e ^ X-1 / x, let t = e ^ X-1

Let t = e ^ X-1, x = ln (T + 1)
Original formula = t / ln (T + 1)
=1/[(1/t)ln(t+1)]
=1/ln(1+t)^(1/t)(t->0)
=1/lne
=1
Solution 2
The original formula = (e ^ x-1) / X (x - > 0) = (e ^ x-1) '/ X' (Robita's law) = e ^ X / 1 = e ^ 0 = 1
All roads lead to Rome

Using Taylor formula to find the limit limx approaching 0 (cosx-e ^ - x ^ 2 / 2) / x ^ 4

When x is close to 0, find the limit of (tan3x) / X