Question

Lim(1-cos(x))/(x*sin(x))=0 делить на 0x стремиться к 0

Answer 1

Ответ:

$\lim_{x \to 0} \frac{1-cosx}{x*sinx}= \lim_{x \to 0} \frac{(1-cosx)(1+cosx)}{x^2(1+cosx)}= \lim_{x \to 0} \frac{sin^2x}{x^2(1+cosx)}=1/2$

Пошаговое объяснение:

Answer 2

Let's evaluate the limit of the expression:

Lim(1-cos(x))/(x*sin(x)) as x approaches 0.

To compute this limit, we can use L'Hôpital's rule, which states that for an indeterminate form of 0/0 or ∞/∞, the limit of the ratio of the derivatives of the numerator and denominator will be the same as the original limit.

Taking the derivatives:

1. Derivative of (1-cos(x)) with respect to x: d/dx(1-cos(x)) = sin(x).
2. Derivative of (xsin(x)) with respect to x: d/dx(xsin(x)) = x*cos(x) + sin(x).

Now, let's apply L'Hôpital's rule:

Lim(1-cos(x))/(xsin(x)) as x approaches 0 = Lim(sin(x)/(xcos(x) + sin(x))) as x approaches 0.

Now, we can substitute x=0:

Lim(sin(0)/(0*cos(0) + sin(0))) = Lim(0/0).

We still have an indeterminate form of 0/0. So, we can apply L'Hôpital's rule once again:

1. Derivative of sin(x) with respect to x: d/dx(sin(x)) = cos(x).
2. Derivative of (xcos(x) + sin(x)) with respect to x: d/dx(xcos(x) + sin(x)) = cos(x) - x*sin(x).

Now, let's apply L'Hôpital's rule again:

Lim(sin(x)/(xcos(x) + sin(x))) as x approaches 0 = Lim(cos(x)/(cos(x) - xsin(x))) as x approaches 0.

Substitute x=0:

Lim(cos(0)/(cos(0) - 0*sin(0))) = Lim(1/1) = 1.

Therefore, the limit of (1-cos(x))/(x*sin(x)) as x approaches 0 is 1.