Let f (x) = arctan1 / (1-x). When x tends to 1 minus and 1 plus, what are the limits of F (x) respectively,

Let f (x) = arctan1 / (1-x). When x tends to 1 minus and 1 plus, what are the limits of F (x) respectively,

x→1-
Then 1-x → + ∞
So arctanx → π / 2
x→1+
Then 1-x → - ∞
So arctanx → - π / 2

Find the limit? Lim2 (x + 1) arctan1 / X
It's better to write all the explanations
X tends to zero

When x tends to zero from the right, 1 / X tends to positive infinity, arctanx tends to π / 2, so at this time
lim2(x+1)arctan1/x=2*π/2=π
When x tends to zero from the left, 1 / X tends to negative infinity, arctanx tends to - π / 2, so at this time
lim2(x+1)arctan1/x=2-*π/2=-π

Does f (x) = 2 ^ (1 / x) have left limit or right limit at x = 0?

The left limit is 0

How can arctan (1 / x ^ 2) find the limit of X approaching 0?

A:
Let 1 / x ^ 2 = t
lim(x→0) arctan(1/x^2)
=lim(t→+∞) arctan(t)
=π/2

Find the limit f (x) = 2 (x + 1) arctan (1 / x) when x approaches zero from the right

The limit of X (SiNx + cosx) / (1 + x) when x approaches infinity, the - 1 / | x | power of e multiplied by arctan (1 / x) + cosx when x approaches 0

When x approaches positive infinity, Lim √ x (SiNx + cosx) / (1 + x) = LIM (SiNx + cosx) / (1 / √ x + √ x)
=Limc / (1 / √ x + √ x) = 0 (- radical 2)

Limit problem: (2x-1) ^ 30 (3x + 2) ^ 20 / (2x + 1) ^ 50
I'm a beginner. I don't know much about it. Does anyone know how to solve it?
Is x →∞

Should be the limit when x tends to infinity?
If so, compare the coefficient of the highest order term, and the result is
2^30*3^20/2^50= 3^20/2^20
You can divide x ^ 50 up and down at the same time, and then x tends to infinity

The limit of [3 - √ (5 + 4x)] / √ X-1 when x tends to 1

lim [3-√(5+4x)]/(√x-1)
=lim （4-4x)/{(√x-1)【3+√(5+4x)】}
=lim 4(1-√x)(1+√x)/[6(√x-1)]
=-4/3

Why is the limit of (1 + x ^ 2) ^ 1 / 3-1 equal to (1 / 3) x ^ 2?
When x tends to zero

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When x tends to infinity, the limit of - KX times of 1 plus 3 / X is equal to - 9 times of E, then what is k

Using the limit of (1 + 1 / x) ^ x equal to e, we can get k = 3