Question

Lim x -> 0 (arctg * 2x)/(2sin x)​

Answer 1

Ответ:

$\boldsymbol{\boxed{\lim_{x \to 0} \frac{\text{arctg}\ 2x}{2 \sin x} =1}}$

Примечание:

Первый замечательный предел:

$\boxed{\lim_{x \to 0} \frac{\sin x}{ x}= 1}$

По следствию из первого замечательно предела:

$\boxed{\lim_{x \to 0} \frac{x}{ \sin x}= 1}$

$\boxed{\lim_{x \to 0} \frac{ \ \text{arctg}\ x}{x }=1}$

Пошаговое объяснение:

$\displaystyle \lim_{x \to 0} \frac{\text{arctg}\ 2x}{2 \sin x} = \lim_{x \to 0} \frac{x \ \text{arctg}\ 2x}{2x \sin x}= \lim_{x \to 0} \frac{ \ \text{arctg}\ 2x}{2x } \cdot \lim_{x \to 0} \frac{x}{ \sin x}= 1 \cdot 1 =1$

Answer 2

To find the limit of the given expression as x approaches 0:

lim (x -> 0) (arctan(2x))/(2sin(x))

We can use L'Hôpital's Rule because it's an indeterminate form (0/0). L'Hôpital's Rule states that if the limit of the ratio of two functions is of the form 0/0 or ∞/∞, then you can take the derivative of the numerator and the derivative of the denominator separately and evaluate the limit again.

So, let's take the derivatives:

Derivative of the numerator (arctan(2x)): d/dx [arctan(2x)] = 1 / (1 + (2x)^2) * 2 = 2 / (4x^2 + 1)

Derivative of the denominator (2sin(x)): d/dx [2sin(x)] = 2cos(x)

Now, let's find the limit again:

lim (x -> 0) (2 / (4x^2 + 1)) / (2cos(x))

At this point, you can directly substitute x = 0 into the expression:

(2 / (4(0)^2 + 1)) / (2cos(0)) (2 / (1)) / (2 * 1) 2 / 2 1

So, the limit as x approaches 0 of (arctan(2x))/(2sin(x)) is 1.