Solved on Jan 25, 2024

Find the limit of $\frac{\sin 2x}{x^2 \cot 4x}$ as $x$ approaches 0.

STEP 1

Assumptions
1. We are to evaluate the limit of the function $\frac{\sin 2x}{x^{2} \cot 4x}$ as $x$ approaches 0.
2. We will use the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .
3. We will use the trigonometric identity $\cot x = \frac{1}{\tan x}$ .
4. We will use the trigonometric identity $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$ .

STEP 2

First, let's rewrite the function using trigonometric identities. We know that $\cot x = \frac{\cos x}{\sin x}$ , so we can write $\cot 4x$ as $\frac{\cos 4x}{\sin 4x}$ .
$\frac{\sin 2x}{x^{2} \cot 4x} = \frac{\sin 2x}{x^{2}} \cdot \frac{\sin 4x}{\cos 4x}$

STEP 3

Now let's simplify the expression by separating the terms involving $x$ and the trigonometric functions.
$\frac{\sin 2x}{x^{2}} \cdot \frac{\sin 4x}{\cos 4x} = \frac{\sin 2x}{x} \cdot \frac{\sin 4x}{x} \cdot \frac{1}{\cos 4x}$

STEP 4

We can now evaluate the limit of each part separately as $x$ approaches 0. For the first term, we use the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ . However, we need to adjust it for $\sin 2x$ .
$\lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{x \to 0} 2 \cdot \frac{\sin 2x}{2x}$

STEP 5

Using the standard limit, we get:
$\lim_{x \to 0} 2 \cdot \frac{\sin 2x}{2x} = 2 \cdot \lim_{x \to 0} \frac{\sin 2x}{2x} = 2 \cdot 1 = 2$

STEP 6

For the second term, we need to adjust the standard limit for $\sin 4x$ .
$\lim_{x \to 0} \frac{\sin 4x}{x} = \lim_{x \to 0} 4 \cdot \frac{\sin 4x}{4x}$

STEP 7

Using the standard limit, we get:
$\lim_{x \to 0} 4 \cdot \frac{\sin 4x}{4x} = 4 \cdot \lim_{x \to 0} \frac{\sin 4x}{4x} = 4 \cdot 1 = 4$

STEP 8

For the third term, $\cos 4x$ , we know that $\lim_{x \to 0} \cos 4x = \cos 0 = 1$ .

STEP 9

Now we can combine the results of the limits.
$\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = \lim_{x \to 0} \frac{\sin 2x}{x} \cdot \lim_{x \to 0} \frac{\sin 4x}{x} \cdot \lim_{x \to 0} \frac{1}{\cos 4x}$

STEP 10

Substitute the values we found for each limit.
$\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = 2 \cdot 4 \cdot 1$

STEP 11

Calculate the product of the limits.
$\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = 2 \cdot 4 \cdot 1 = 8$
The correct answer is d. 8.

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Find the limit of $\frac{\sin 2x}{x^2 \cot 4x}$ as $x$ approaches 0.

STEP 1

STEP 2

First, let's rewrite the function using trigonometric identities. We know that $\cot x = \frac{\cos x}{\sin x}$ , so we can write $\cot 4x$ as $\frac{\cos 4x}{\sin 4x}$ .
$\frac{\sin 2x}{x^{2} \cot 4x} = \frac{\sin 2x}{x^{2}} \cdot \frac{\sin 4x}{\cos 4x}$

STEP 3

Now let's simplify the expression by separating the terms involving $x$ and the trigonometric functions.
$\frac{\sin 2x}{x^{2}} \cdot \frac{\sin 4x}{\cos 4x} = \frac{\sin 2x}{x} \cdot \frac{\sin 4x}{x} \cdot \frac{1}{\cos 4x}$

STEP 4

We can now evaluate the limit of each part separately as $x$ approaches 0. For the first term, we use the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ . However, we need to adjust it for $\sin 2x$ .
$\lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{x \to 0} 2 \cdot \frac{\sin 2x}{2x}$

STEP 5

Using the standard limit, we get:
$\lim_{x \to 0} 2 \cdot \frac{\sin 2x}{2x} = 2 \cdot \lim_{x \to 0} \frac{\sin 2x}{2x} = 2 \cdot 1 = 2$

STEP 6

For the second term, we need to adjust the standard limit for $\sin 4x$ .
$\lim_{x \to 0} \frac{\sin 4x}{x} = \lim_{x \to 0} 4 \cdot \frac{\sin 4x}{4x}$

STEP 7

Using the standard limit, we get:
$\lim_{x \to 0} 4 \cdot \frac{\sin 4x}{4x} = 4 \cdot \lim_{x \to 0} \frac{\sin 4x}{4x} = 4 \cdot 1 = 4$

STEP 8

For the third term, $\cos 4x$ , we know that $\lim_{x \to 0} \cos 4x = \cos 0 = 1$ .

STEP 9

Now we can combine the results of the limits.
$\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = \lim_{x \to 0} \frac{\sin 2x}{x} \cdot \lim_{x \to 0} \frac{\sin 4x}{x} \cdot \lim_{x \to 0} \frac{1}{\cos 4x}$

STEP 10

Substitute the values we found for each limit.
$\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = 2 \cdot 4 \cdot 1$

STEP 11

Calculate the product of the limits.
$\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = 2 \cdot 4 \cdot 1 = 8$
The correct answer is d. 8.

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Find the limit of sin⁡2xx2cot⁡4x\frac{\sin 2x}{x^2 \cot 4x}x2cot4xsin2x​ as xxx approaches 0.

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

Using the standard limit, we get:lim⁡x→02⋅sin⁡2x2x=2⋅lim⁡x→0sin⁡2x2x=2⋅1=2\lim_{x \to 0} 2 \cdot \frac{\sin 2x}{2x} = 2 \cdot \lim_{x \to 0} \frac{\sin 2x}{2x} = 2 \cdot 1 = 2x→0lim​2⋅2xsin2x​=2⋅x→0lim​2xsin2x​=2⋅1=2

STEP 6

For the second term, we need to adjust the standard limit for sin⁡4x\sin 4xsin4x.lim⁡x→0sin⁡4xx=lim⁡x→04⋅sin⁡4x4x\lim_{x \to 0} \frac{\sin 4x}{x} = \lim_{x \to 0} 4 \cdot \frac{\sin 4x}{4x}x→0lim​xsin4x​=x→0lim​4⋅4xsin4x​

STEP 7

Using the standard limit, we get:lim⁡x→04⋅sin⁡4x4x=4⋅lim⁡x→0sin⁡4x4x=4⋅1=4\lim_{x \to 0} 4 \cdot \frac{\sin 4x}{4x} = 4 \cdot \lim_{x \to 0} \frac{\sin 4x}{4x} = 4 \cdot 1 = 4x→0lim​4⋅4xsin4x​=4⋅x→0lim​4xsin4x​=4⋅1=4

STEP 8

For the third term, cos⁡4x\cos 4xcos4x, we know that lim⁡x→0cos⁡4x=cos⁡0=1\lim_{x \to 0} \cos 4x = \cos 0 = 1limx→0​cos4x=cos0=1.

STEP 9

STEP 10

Substitute the values we found for each limit.lim⁡x→0sin⁡2xx2cot⁡4x=2⋅4⋅1\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = 2 \cdot 4 \cdot 1x→0lim​x2cot4xsin2x​=2⋅4⋅1

STEP 11

Calculate the product of the limits.lim⁡x→0sin⁡2xx2cot⁡4x=2⋅4⋅1=8\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = 2 \cdot 4 \cdot 1 = 8x→0lim​x2cot4xsin2x​=2⋅4⋅1=8The correct answer is d. 8.

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Find the limit of $\frac{\sin 2x}{x^2 \cot 4x}$ as $x$ approaches 0.

Using the standard limit, we get:
$\lim_{x \to 0} 2 \cdot \frac{\sin 2x}{2x} = 2 \cdot \lim_{x \to 0} \frac{\sin 2x}{2x} = 2 \cdot 1 = 2$

For the second term, we need to adjust the standard limit for $\sin 4x$ .
$\lim_{x \to 0} \frac{\sin 4x}{x} = \lim_{x \to 0} 4 \cdot \frac{\sin 4x}{4x}$

Using the standard limit, we get:
$\lim_{x \to 0} 4 \cdot \frac{\sin 4x}{4x} = 4 \cdot \lim_{x \to 0} \frac{\sin 4x}{4x} = 4 \cdot 1 = 4$

For the third term, $\cos 4x$ , we know that $\lim_{x \to 0} \cos 4x = \cos 0 = 1$ .

Substitute the values we found for each limit.
$\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = 2 \cdot 4 \cdot 1$

Calculate the product of the limits.
$\lim_{x \to 0} \frac{\sin 2x}{x^{2} \cot 4x} = 2 \cdot 4 \cdot 1 = 8$
The correct answer is d. 8.