Question

Limxстремится 0 1-cos3x/x^2

Answer 1

Ответ:1/e^3

Пошаговое объяснение:1)(2 - √(6 + x) / √(7 - x)) - 3x = (2 - √(6 - 2) / √(7 + 2)) + 6 = (2 - √4)/√9 + 6 = 0/3 + 6 = 6

2)sin^(-1)(5x) * cos(2x) / sin(2x) = cos(2x) * 5 / √(1 - 25x^2) / 2 * cos(2x) = cos (2*0) * 5 / √(1 - 25* 0^2) / 2 * cos(2 * 0) = 5/2

3)((x - 5) / (x - 2))^x = exp (log ((x - 5) / (x - 2))^x) = exp(x * log((x - 5) / (x - 2))) = exp(-3x^2 / ((x - 5) / (x - 2))) = exp (-3 * 2 * x/(2x - 7)) = e^(-3*2/2) = 1/e^3

Answer 2

To find the limit of the function as x approaches 0, we can use L'Hôpital's rule, which allows us to evaluate limits of indeterminate forms (0/0 or infinity/infinity) by taking the derivative of the numerator and denominator separately.

The given function is:

f(x) = (1 - cos(3x)) / x^2

As x approaches 0, the function is in the indeterminate form 0/0. Let's find the derivatives of the numerator and denominator to apply L'Hôpital's rule:

f'(x) = d/dx(1 - cos(3x)) = 3sin(3x)

g'(x) = d/dx(x^2) = 2x

Now, we can take the limit of the derivatives as x approaches 0:

lim (x → 0) [f'(x) / g'(x)] = lim (x → 0) [(3sin(3x)) / (2x)]

Now, plug in x = 0 to evaluate the limit:

lim (x → 0) [(3sin(3x)) / (2x)] = (3sin(3 * 0)) / (2 * 0) = 0 / 0

Once again, we get an indeterminate form (0/0). We can apply L'Hôpital's rule again:

f''(x) = d/dx(3sin(3x)) = 9cos(3x)

g''(x) = d/dx(2x) = 2

Now, let's take the limit of the second derivatives as x approaches 0:

lim (x → 0) [f''(x) / g''(x)] = lim (x → 0) [(9cos(3x)) / 2]

Now, plug in x = 0 to evaluate the limit:

lim (x → 0) [(9cos(3x)) / 2] = (9cos(3 * 0)) / 2 = 9 / 2

Finally, the limit of the original function as x approaches 0 is:

lim (x → 0) [(1 - cos(3x)) / x^2] = 9 / 2