Question

Lim x>0 ((2sin^2 x/2)/x^2)

Answer 1

Преобразуем выражение, стоящее под знаком предела:

Числитель:

2*sin (x/2)*sin(x/2)

Знаменатель:
4*(x/2)*(x/2)

Используем замечательный предел: При x→0  lim ( sin(x/2) / (x/2) ) = 1

Получаем:

2*sin(x/2)*sin(x/2) / (4*(x/2)*(x/2))

2*1*1 / 4 = 1/2

Ответ: 1/2

Проверим наше решение программно:
См. скрин:

Ответы совпали, значит наше решение - правильное!

Answer 2

To evaluate the limit of the given expression, let's break it down step by step.

First, simplify the expression: (2sin^2(x/2)) / x^2

Next, apply the double-angle identity for sine: sin^2(x/2) = (1 - cos(x)) / 2

Now, substitute this back into the expression: (2(1 - cos(x))/2) / x^2 = (1 - cos(x)) / x^2

As x approaches 0, cos(x) approaches 1. Therefore, in the limit as x approaches 0, the expression becomes: (1 - 1) / 0^2 = 0 / 0

This is an indeterminate form. We can apply L'Hôpital's rule to find the limit. Taking the derivative of the numerator and denominator separately:

Numerator: d/dx (1 - cos(x)) = sin(x)

Denominator: d/dx (x^2) = 2x

Now, apply L'Hôpital's rule: lim x->0 (sin(x)) / (2x)

At x = 0, sin(x) = 0 and 2x = 0, so we can directly substitute the values: lim x->0 (0) / (0) = 0

Therefore, the limit of the given expression as x approaches 0 is 0.