Question

ВЫРУЧАЙТЕ!!! СРОЧНО НАДО!!! ПОЖАЛУЙСТА!!! Найти предел функций, не пользуюсь правило Лопиталя: Lim(x->0) (arcsin 3x)/(5x)

Answer 1

Извествно что lim(sinx/x)=1 значит lim(x/arcsinx)=1, значит arcsin3x можно  заменить на 3x
Получаем lim(x->0) (3x/5x)=3/5

Answer 2

Finding the Limit of a Function without Using L'Hôpital's Rule

To find the limit of the function lim(x->0) (arcsin 3x)/(5x) without using L'Hôpital's Rule, we can apply some trigonometric identities and properties of limits.

Trigonometric Identity:

The arcsin function can be expressed in terms of sine function using the identity arcsin(x) = sin^(-1)(x). This allows us to rewrite the given function in terms of sine function.

Applying the Trigonometric Identity:

The given function lim(x->0) (arcsin 3x)/(5x) can be rewritten as lim(x->0) (sin^(-1)(3x))/(5x).

Using the Property of Limits:

We know that lim(x->0) sin(x)/x = 1. We can use this property to simplify the given function.

Simplifying the Function:

By applying the property lim(x->0) sin(x)/x = 1, we can rewrite the function as (1/3) * lim(x->0) (sin^(-1)(3x))/(3x).

Evaluating the Limit:

Now, we can evaluate the limit lim(x->0) (sin^(-1)(3x))/(3x). As x approaches 0, the expression (sin^(-1)(3x))/(3x) approaches 1/3.

Therefore, the limit of the given function lim(x->0) (arcsin 3x)/(5x) without using L'Hôpital's Rule is 1/3.

I hope this helps! If you have any further questions or need additional assistance, feel free to ask.