Calculus Examples

$lim_{x \to 0} \frac{2 \sin (x) - \sin (2 x)}{x - \sin (x)}$

Step 1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to 0} 2 \sin (x) - \sin (2 x)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{lim_{x \to 0} 2 \sin (x) - lim_{x \to 0} \sin (2 x)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.2

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{2 lim_{x \to 0} \sin (x) - lim_{x \to 0} \sin (2 x)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.3

Move the limit inside the trig function because sine is continuous.

$\frac{2 \sin (lim_{x \to 0} x) - lim_{x \to 0} \sin (2 x)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.4

Move the limit inside the trig function because sine is continuous.

$\frac{2 \sin (lim_{x \to 0} x) - \sin (lim_{x \to 0} 2 x)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.5

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{2 \sin (lim_{x \to 0} x) - \sin (2 lim_{x \to 0} x)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.6

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 1.2.6.1

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{2 \sin (0) - \sin (2 lim_{x \to 0} x)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.6.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{2 \sin (0) - \sin (2 \cdot 0)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.7

Simplify the answer.

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Step 1.2.7.1

Simplify each term.

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Step 1.2.7.1.1

The exact value of $\sin (0)$ is $0$ .

$\frac{2 \cdot 0 - \sin (2 \cdot 0)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.7.1.2

Multiply $2$ by $0$ .

$\frac{0 - \sin (2 \cdot 0)}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.7.1.3

Multiply $2$ by $0$ .

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

Step 1.2.7.1.4

The exact value of $\sin (0)$ is $0$ .

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

Step 1.2.7.1.5

Multiply $- 1$ by $0$ .

$\frac{0 + 0}{lim_{x \to 0} x - \sin (x)}$

Step 1.2.7.2

Add $0$ and $0$ .

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

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

Step 1.3.2

Move the limit inside the trig function because sine is continuous.

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

Step 1.3.3

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 1.3.3.1

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

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

Step 1.3.3.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{0}{0 - \sin (0)}$

Step 1.3.4

Simplify the answer.

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Step 1.3.4.1

Simplify each term.

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Step 1.3.4.1.1

The exact value of $\sin (0)$ is $0$ .

$\frac{0}{0 - 0}$

Step 1.3.4.1.2

Multiply $- 1$ by $0$ .

$\frac{0}{0 + 0}$

Step 1.3.4.2

Add $0$ and $0$ .

$\frac{0}{0}$

Step 1.3.4.3

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.3.5

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 0} \frac{2 \sin (x) - \sin (2 x)}{x - \sin (x)} = lim_{x \to 0} \frac{\frac{d}{d x} [2 \sin (x) - \sin (2 x)]}{\frac{d}{d x} [x - \sin (x)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [2 \sin (x) - \sin (2 x)]}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.2

By the Sum Rule, the derivative of $2 \sin (x) - \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [2 \sin (x)] + \frac{d}{d x} [- \sin (2 x)]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [2 \sin (x)] + \frac{d}{d x} [- \sin (2 x)]}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.3

Evaluate $\frac{d}{d x} [2 \sin (x)]$ .

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Step 3.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 \sin (x)$ with respect to $x$ is $2 \frac{d}{d x} [\sin (x)]$ .

$lim_{x \to 0} \frac{2 \frac{d}{d x} [\sin (x)] + \frac{d}{d x} [- \sin (2 x)]}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.3.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$lim_{x \to 0} \frac{2 \cos (x) + \frac{d}{d x} [- \sin (2 x)]}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4

Evaluate $\frac{d}{d x} [- \sin (2 x)]$ .

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Step 3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (2 x)$ with respect to $x$ is $- \frac{d}{d x} [\sin (2 x)]$ .

$lim_{x \to 0} \frac{2 \cos (x) - \frac{d}{d x} [\sin (2 x)]}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 3.4.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to 0} \frac{2 \cos (x) - (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x])}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4.2.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$lim_{x \to 0} \frac{2 \cos (x) - (\cos (u) \frac{d}{d x} [2 x])}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4.2.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to 0} \frac{2 \cos (x) - (\cos (2 x) \frac{d}{d x} [2 x])}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{2 \cos (x) - (\cos (2 x) (2 \frac{d}{d x} [x]))}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 0} \frac{2 \cos (x) - (\cos (2 x) (2 \cdot 1))}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4.5

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{2 \cos (x) - (\cos (2 x) \cdot 2)}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4.6

Move $2$ to the left of $\cos (2 x)$ .

$lim_{x \to 0} \frac{2 \cos (x) - (2 \cos (2 x))}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.4.7

Multiply $2$ by $- 1$ .

$lim_{x \to 0} \frac{2 \cos (x) - 2 \cos (2 x)}{\frac{d}{d x} [x - \sin (x)]}$

Step 3.5

By the Sum Rule, the derivative of $x - \sin (x)$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- \sin (x)]$ .

$lim_{x \to 0} \frac{2 \cos (x) - 2 \cos (2 x)}{\frac{d}{d x} [x] + \frac{d}{d x} [- \sin (x)]}$

Step 3.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 0} \frac{2 \cos (x) - 2 \cos (2 x)}{1 + \frac{d}{d x} [- \sin (x)]}$

Step 3.7

Evaluate $\frac{d}{d x} [- \sin (x)]$ .

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Step 3.7.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (x)$ with respect to $x$ is $- \frac{d}{d x} [\sin (x)]$ .

$lim_{x \to 0} \frac{2 \cos (x) - 2 \cos (2 x)}{1 - \frac{d}{d x} [\sin (x)]}$

Step 3.7.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$lim_{x \to 0} \frac{2 \cos (x) - 2 \cos (2 x)}{1 - \cos (x)}$

Step 4

Apply L'Hospital's rule.

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Step 4.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 4.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to 0} 2 \cos (x) - 2 \cos (2 x)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2

Evaluate the limit of the numerator.

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Step 4.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{lim_{x \to 0} 2 \cos (x) - lim_{x \to 0} 2 \cos (2 x)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.2

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{2 lim_{x \to 0} \cos (x) - lim_{x \to 0} 2 \cos (2 x)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.3

Move the limit inside the trig function because cosine is continuous.

$\frac{2 \cos (lim_{x \to 0} x) - lim_{x \to 0} 2 \cos (2 x)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{2 \cos (lim_{x \to 0} x) - 2 lim_{x \to 0} \cos (2 x)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.5

Move the limit inside the trig function because cosine is continuous.

$\frac{2 \cos (lim_{x \to 0} x) - 2 \cos (lim_{x \to 0} 2 x)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.6

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{2 \cos (lim_{x \to 0} x) - 2 \cos (2 lim_{x \to 0} x)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.7

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 4.1.2.7.1

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{2 \cos (0) - 2 \cos (2 lim_{x \to 0} x)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.7.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{2 \cos (0) - 2 \cos (2 \cdot 0)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.8

Simplify the answer.

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Step 4.1.2.8.1

Simplify each term.

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Step 4.1.2.8.1.1

The exact value of $\cos (0)$ is $1$ .

$\frac{2 \cdot 1 - 2 \cos (2 \cdot 0)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.8.1.2

Multiply $2$ by $1$ .

$\frac{2 - 2 \cos (2 \cdot 0)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.8.1.3

Multiply $2$ by $0$ .

$\frac{2 - 2 \cos (0)}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.8.1.4

The exact value of $\cos (0)$ is $1$ .

$\frac{2 - 2 \cdot 1}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.8.1.5

Multiply $- 2$ by $1$ .

$\frac{2 - 2}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.2.8.2

Subtract $2$ from $2$ .

$\frac{0}{lim_{x \to 0} 1 - \cos (x)}$

Step 4.1.3

Evaluate the limit of the denominator.

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Step 4.1.3.1

Evaluate the limit.

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Step 4.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{0}{lim_{x \to 0} 1 - lim_{x \to 0} \cos (x)}$

Step 4.1.3.1.2

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

$\frac{0}{1 - lim_{x \to 0} \cos (x)}$

Step 4.1.3.1.3

Move the limit inside the trig function because cosine is continuous.

$\frac{0}{1 - \cos (lim_{x \to 0} x)}$

Step 4.1.3.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{0}{1 - \cos (0)}$

Step 4.1.3.3

Simplify the answer.

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Step 4.1.3.3.1

Simplify each term.

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Step 4.1.3.3.1.1

The exact value of $\cos (0)$ is $1$ .

$\frac{0}{1 - 1 \cdot 1}$

Step 4.1.3.3.1.2

Multiply $- 1$ by $1$ .

$\frac{0}{1 - 1}$

Step 4.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 4.1.3.3.3

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 4.1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 4.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 4.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 0} \frac{2 \cos (x) - 2 \cos (2 x)}{1 - \cos (x)} = lim_{x \to 0} \frac{\frac{d}{d x} [2 \cos (x) - 2 \cos (2 x)]}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [2 \cos (x) - 2 \cos (2 x)]}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.2

By the Sum Rule, the derivative of $2 \cos (x) - 2 \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [2 \cos (x)] + \frac{d}{d x} [- 2 \cos (2 x)]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [2 \cos (x)] + \frac{d}{d x} [- 2 \cos (2 x)]}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.3

Evaluate $\frac{d}{d x} [2 \cos (x)]$ .

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Step 4.3.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 \cos (x)$ with respect to $x$ is $2 \frac{d}{d x} [\cos (x)]$ .

$lim_{x \to 0} \frac{2 \frac{d}{d x} [\cos (x)] + \frac{d}{d x} [- 2 \cos (2 x)]}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.3.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$lim_{x \to 0} \frac{2 (- \sin (x)) + \frac{d}{d x} [- 2 \cos (2 x)]}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.3.3

Multiply $- 1$ by $2$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + \frac{d}{d x} [- 2 \cos (2 x)]}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4

Evaluate $\frac{d}{d x} [- 2 \cos (2 x)]$ .

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Step 4.3.4.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 \cos (2 x)$ with respect to $x$ is $- 2 \frac{d}{d x} [\cos (2 x)]$ .

$lim_{x \to 0} \frac{- 2 \sin (x) - 2 \frac{d}{d x} [\cos (2 x)]}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 4.3.4.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to 0} \frac{- 2 \sin (x) - 2 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x])}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4.2.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$lim_{x \to 0} \frac{- 2 \sin (x) - 2 (- \sin (u) \frac{d}{d x} [2 x])}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4.2.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to 0} \frac{- 2 \sin (x) - 2 (- \sin (2 x) \frac{d}{d x} [2 x])}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{- 2 \sin (x) - 2 (- \sin (2 x) (2 \frac{d}{d x} [x]))}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 0} \frac{- 2 \sin (x) - 2 (- \sin (2 x) (2 \cdot 1))}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4.5

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{- 2 \sin (x) - 2 (- \sin (2 x) \cdot 2)}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4.6

Multiply $2$ by $- 1$ .

$lim_{x \to 0} \frac{- 2 \sin (x) - 2 (- 2 \sin (2 x))}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.4.7

Multiply $- 2$ by $- 2$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{\frac{d}{d x} [1 - \cos (x)]}$

Step 4.3.5

By the Sum Rule, the derivative of $1 - \cos (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (x)]$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (x)]}$

Step 4.3.6

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{0 + \frac{d}{d x} [- \cos (x)]}$

Step 4.3.7

Evaluate $\frac{d}{d x} [- \cos (x)]$ .

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Step 4.3.7.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (x)]$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{0 - \frac{d}{d x} [\cos (x)]}$

Step 4.3.7.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{0 - - \sin (x)}$

Step 4.3.7.3

Multiply $- 1$ by $- 1$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{0 + 1 \sin (x)}$

Step 4.3.7.4

Multiply $\sin (x)$ by $1$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{0 + \sin (x)}$

Step 4.3.8

Add $0$ and $\sin (x)$ .

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{\sin (x)}$

Step 5

Apply L'Hospital's rule.

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Step 5.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 5.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to 0} - 2 \sin (x) + 4 \sin (2 x)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2

Evaluate the limit of the numerator.

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Step 5.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{- lim_{x \to 0} 2 \sin (x) + lim_{x \to 0} 4 \sin (2 x)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.2

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{- 2 lim_{x \to 0} \sin (x) + lim_{x \to 0} 4 \sin (2 x)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.3

Move the limit inside the trig function because sine is continuous.

$\frac{- 2 \sin (lim_{x \to 0} x) + lim_{x \to 0} 4 \sin (2 x)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.4

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{- 2 \sin (lim_{x \to 0} x) + 4 lim_{x \to 0} \sin (2 x)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.5

Move the limit inside the trig function because sine is continuous.

$\frac{- 2 \sin (lim_{x \to 0} x) + 4 \sin (lim_{x \to 0} 2 x)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.6

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{- 2 \sin (lim_{x \to 0} x) + 4 \sin (2 lim_{x \to 0} x)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.7

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 5.1.2.7.1

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{- 2 \sin (0) + 4 \sin (2 lim_{x \to 0} x)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.7.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{- 2 \sin (0) + 4 \sin (2 \cdot 0)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.8

Simplify the answer.

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Step 5.1.2.8.1

Simplify each term.

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Step 5.1.2.8.1.1

The exact value of $\sin (0)$ is $0$ .

$\frac{- 2 \cdot 0 + 4 \sin (2 \cdot 0)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.8.1.2

Multiply $- 2$ by $0$ .

$\frac{0 + 4 \sin (2 \cdot 0)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.8.1.3

Multiply $2$ by $0$ .

$\frac{0 + 4 \sin (0)}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.8.1.4

The exact value of $\sin (0)$ is $0$ .

$\frac{0 + 4 \cdot 0}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.8.1.5

Multiply $4$ by $0$ .

$\frac{0 + 0}{lim_{x \to 0} \sin (x)}$

Step 5.1.2.8.2

Add $0$ and $0$ .

$\frac{0}{lim_{x \to 0} \sin (x)}$

Step 5.1.3

Evaluate the limit of the denominator.

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Step 5.1.3.1

Move the limit inside the trig function because sine is continuous.

$\frac{0}{\sin (lim_{x \to 0} x)}$

Step 5.1.3.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{0}{\sin (0)}$

Step 5.1.3.3

The exact value of $\sin (0)$ is $0$ .

$\frac{0}{0}$

Step 5.1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 5.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 5.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 0} \frac{- 2 \sin (x) + 4 \sin (2 x)}{\sin (x)} = lim_{x \to 0} \frac{\frac{d}{d x} [- 2 \sin (x) + 4 \sin (2 x)]}{\frac{d}{d x} [\sin (x)]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [- 2 \sin (x) + 4 \sin (2 x)]}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.2

By the Sum Rule, the derivative of $- 2 \sin (x) + 4 \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [- 2 \sin (x)] + \frac{d}{d x} [4 \sin (2 x)]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [- 2 \sin (x)] + \frac{d}{d x} [4 \sin (2 x)]}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.3

Evaluate $\frac{d}{d x} [- 2 \sin (x)]$ .

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Step 5.3.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 \sin (x)$ with respect to $x$ is $- 2 \frac{d}{d x} [\sin (x)]$ .

$lim_{x \to 0} \frac{- 2 \frac{d}{d x} [\sin (x)] + \frac{d}{d x} [4 \sin (2 x)]}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.3.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + \frac{d}{d x} [4 \sin (2 x)]}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4

Evaluate $\frac{d}{d x} [4 \sin (2 x)]$ .

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Step 5.3.4.1

Since $4$ is constant with respect to $x$ , the derivative of $4 \sin (2 x)$ with respect to $x$ is $4 \frac{d}{d x} [\sin (2 x)]$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 4 \frac{d}{d x} [\sin (2 x)]}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 5.3.4.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 4 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x])}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4.2.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 4 (\cos (u) \frac{d}{d x} [2 x])}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4.2.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 4 (\cos (2 x) \frac{d}{d x} [2 x])}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 4 (\cos (2 x) (2 \frac{d}{d x} [x]))}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 4 (\cos (2 x) (2 \cdot 1))}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4.5

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 4 (\cos (2 x) \cdot 2)}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4.6

Move $2$ to the left of $\cos (2 x)$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 4 (2 \cdot \cos (2 x))}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.4.7

Multiply $2$ by $4$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 8 \cos (2 x)}{\frac{d}{d x} [\sin (x)]}$

Step 5.3.5

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$lim_{x \to 0} \frac{- 2 \cos (x) + 8 \cos (2 x)}{\cos (x)}$

Step 6

Evaluate the limit.

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Step 6.1

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $0$ .

$\frac{lim_{x \to 0} - 2 \cos (x) + 8 \cos (2 x)}{lim_{x \to 0} \cos (x)}$

Step 6.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{- lim_{x \to 0} 2 \cos (x) + lim_{x \to 0} 8 \cos (2 x)}{lim_{x \to 0} \cos (x)}$

Step 6.3

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{- 2 lim_{x \to 0} \cos (x) + lim_{x \to 0} 8 \cos (2 x)}{lim_{x \to 0} \cos (x)}$

Step 6.4

Move the limit inside the trig function because cosine is continuous.

$\frac{- 2 \cos (lim_{x \to 0} x) + lim_{x \to 0} 8 \cos (2 x)}{lim_{x \to 0} \cos (x)}$

Step 6.5

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$\frac{- 2 \cos (lim_{x \to 0} x) + 8 lim_{x \to 0} \cos (2 x)}{lim_{x \to 0} \cos (x)}$

Step 6.6

Move the limit inside the trig function because cosine is continuous.

$\frac{- 2 \cos (lim_{x \to 0} x) + 8 \cos (lim_{x \to 0} 2 x)}{lim_{x \to 0} \cos (x)}$

Step 6.7

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{- 2 \cos (lim_{x \to 0} x) + 8 \cos (2 lim_{x \to 0} x)}{lim_{x \to 0} \cos (x)}$

Step 6.8

Move the limit inside the trig function because cosine is continuous.

$\frac{- 2 \cos (lim_{x \to 0} x) + 8 \cos (2 lim_{x \to 0} x)}{\cos (lim_{x \to 0} x)}$

Step 7

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 7.1

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{- 2 \cos (0) + 8 \cos (2 lim_{x \to 0} x)}{\cos (lim_{x \to 0} x)}$

Step 7.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{- 2 \cos (0) + 8 \cos (2 \cdot 0)}{\cos (lim_{x \to 0} x)}$

Step 7.3

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{- 2 \cos (0) + 8 \cos (2 \cdot 0)}{\cos (0)}$

Step 8

Simplify the answer.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

The exact value of $\cos (0)$ is $1$ .

$\frac{- 2 \cdot 1 + 8 \cos (2 \cdot 0)}{\cos (0)}$

Step 8.1.2

Multiply $- 2$ by $1$ .

$\frac{- 2 + 8 \cos (2 \cdot 0)}{\cos (0)}$

Step 8.1.3

Multiply $2$ by $0$ .

$\frac{- 2 + 8 \cos (0)}{\cos (0)}$

Step 8.1.4

The exact value of $\cos (0)$ is $1$ .

$\frac{- 2 + 8 \cdot 1}{\cos (0)}$

Step 8.1.5

Multiply $8$ by $1$ .

$\frac{- 2 + 8}{\cos (0)}$

Step 8.1.6

Add $- 2$ and $8$ .

$\frac{6}{\cos (0)}$

Step 8.2

The exact value of $\cos (0)$ is $1$ .

$\frac{6}{1}$

Step 8.3

Divide $6$ by $1$ .

$6$

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