Question

Lim x-0 (tgx-sinx)/x^3

Answer 1

Ответ:

Пошаговое объяснение: lim(x→0) ((tgx-sinx)/x³)=1/2.

lim(x→0) ((tgx-sinx)/x³)

Неопределённлсть 0/0.   ⇒

Нужно брать одновременно призводные от числителя и знаменателя,

причём знаменатель должен быть числом, отличным от нуля:

(x³)'=3x²

(3x²)'=6x

(6x)'=6.  ⇒

От числителя нужно взять три производных.

Так как в числителе сумма, ⇒ берём производные по отдельности:

1.

(tgx)'=1/cos²x=(cosx)⁻²

((cosx)⁻²)'=-2*(cosx)⁻³*(-sinx)=2*(cosx)⁻³*sinx

(2*(cosx)⁻³*sinx)'=-6*(cosx)⁻⁴*sinx+2*(cosx)⁻²*cosx=

=-6*(cosx)⁻⁴*sinx+(2/(cosx)).

2.

(-sinx)'=-cosx

(-cosx)'=sinx

(sinx)'=cosx    ⇒

(-6*(cos0)⁻⁴*sin0+(2/(cos0))+cos0)/6=(-6*1*0+(2/1)+1)/6=3/6=1/2.

Answer 2

To evaluate the limit of the function $\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3}$, we can use L'Hôpital's rule, which states that if the limit of the ratio of two functions exists in an indeterminate form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then the limit of their derivatives will be the same.

Let's apply L'Hôpital's rule to the given function:

$\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3}$

We can differentiate the numerator and the denominator separately:

Numerator: $\frac{d}{dx}(\tan(x) - \sin(x)) = \sec^2(x) - \cos(x)$

Denominator: $\frac{d}{dx}(x^3) = 3x^2$

Now we can rewrite the limit using the derivatives:

$\lim_{x \to 0} \frac{\sec^2(x) - \cos(x)}{3x^2}$

Let's evaluate the limit again at $x = 0$:

$\frac{\sec^2(0) - \cos(0)}{3(0)^2} = \frac{1 - 1}{0} = \frac{0}{0}$

We still have an indeterminate form. Applying L'Hôpital's rule again:

Numerator: $\frac{d}{dx}(\sec^2(x) - \cos(x)) = 2\sec(x)\tan(x) + \sin(x)$

Denominator: $\frac{d}{dx}(3x^2) = 6x$

Now we can rewrite the limit using the new derivatives:

$\lim_{x \to 0} \frac{2\sec(x)\tan(x) + \sin(x)}{6x}$

Evaluating the limit at $x = 0$ once more:

$\frac{2\sec(0)\tan(0) + \sin(0)}{6(0)} = \frac{0}{0}$

We still have an indeterminate form. Applying L'Hôpital's rule for the third time:

Numerator: $\frac{d}{dx}(2\sec(x)\tan(x) + \sin(x)) = 2\sec^2(x) + 2\sec(x)\tan^2(x) + \cos(x)$

Denominator: $\frac{d}{dx}(6x) = 6$

Now we can rewrite the limit using the new derivatives:

$\lim_{x \to 0} \frac{2\sec^2(x) + 2\sec(x)\tan^2(x) + \cos(x)}{6}$

Evaluating the limit at $x = 0$ for the final time:

$\frac{2\sec^2(0) + 2\sec(0)\tan^2(0) + \cos(0)}{6} = \frac{2(1) + 2(1)(0)^2 + 1}{6} = \frac{3}{6} = \frac{1}{2}$

Therefore, the limit of $\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3}$