Calculus Examples

$lim_{x \to 0} \frac{6 \sin (2 x) - 4 x}{x}$

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

$\frac{lim_{x \to 0} 6 \sin (2 x) - 4 x}{lim_{x \to 0} x}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

$\frac{lim_{x \to 0} 6 \sin (2 x) - lim_{x \to 0} 4 x}{lim_{x \to 0} x}$

Step 1.1.2.2

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$\frac{6 lim_{x \to 0} \sin (2 x) - lim_{x \to 0} 4 x}{lim_{x \to 0} x}$

Step 1.1.2.3

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

$\frac{6 \sin (lim_{x \to 0} 2 x) - lim_{x \to 0} 4 x}{lim_{x \to 0} x}$

Step 1.1.2.4

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

$\frac{6 \sin (2 lim_{x \to 0} x) - lim_{x \to 0} 4 x}{lim_{x \to 0} x}$

Step 1.1.2.5

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

$\frac{6 \sin (2 lim_{x \to 0} x) - 4 lim_{x \to 0} x}{lim_{x \to 0} x}$

Step 1.1.2.6

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

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Step 1.1.2.6.1

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

$\frac{6 \sin (2 \cdot 0) - 4 lim_{x \to 0} x}{lim_{x \to 0} x}$

Step 1.1.2.6.2

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

$\frac{6 \sin (2 \cdot 0) - 4 \cdot 0}{lim_{x \to 0} x}$

Step 1.1.2.7

Simplify the answer.

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Step 1.1.2.7.1

Simplify each term.

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Step 1.1.2.7.1.1

Multiply $2$ by $0$ .

$\frac{6 \sin (0) - 4 \cdot 0}{lim_{x \to 0} x}$

Step 1.1.2.7.1.2

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

$\frac{6 \cdot 0 - 4 \cdot 0}{lim_{x \to 0} x}$

Step 1.1.2.7.1.3

Multiply $6$ by $0$ .

$\frac{0 - 4 \cdot 0}{lim_{x \to 0} x}$

Step 1.1.2.7.1.4

Multiply $- 4$ by $0$ .

$\frac{0 + 0}{lim_{x \to 0} x}$

Step 1.1.2.7.2

Add $0$ and $0$ .

$\frac{0}{lim_{x \to 0} x}$

Step 1.1.3

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

$\frac{0}{0}$

Step 1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.2

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.

$lim_{x \to 0} \frac{6 \sin (2 x) - 4 x}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [6 \sin (2 x) - 4 x]}{\frac{d}{d x} [x]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [6 \sin (2 x) - 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.2

By the Sum Rule, the derivative of $6 \sin (2 x) - 4 x$ with respect to $x$ is $\frac{d}{d x} [6 \sin (2 x)] + \frac{d}{d x} [- 4 x]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [6 \sin (2 x)] + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [6 \sin (2 x)]$ .

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Step 1.3.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \sin (2 x)$ with respect to $x$ is $6 \frac{d}{d x} [\sin (2 x)]$ .

$lim_{x \to 0} \frac{6 \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3.2

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 1.3.3.2.1

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

$lim_{x \to 0} \frac{6 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3.2.2

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

$lim_{x \to 0} \frac{6 (\cos (u) \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3.2.3

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

$lim_{x \to 0} \frac{6 (\cos (2 x) \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3.3

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

$lim_{x \to 0} \frac{6 (\cos (2 x) (2 \frac{d}{d x} [x])) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3.4

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

$lim_{x \to 0} \frac{6 (\cos (2 x) (2 \cdot 1)) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3.5

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{6 (\cos (2 x) \cdot 2) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3.6

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

$lim_{x \to 0} \frac{6 (2 \cdot \cos (2 x)) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.3.7

Multiply $2$ by $6$ .

$lim_{x \to 0} \frac{12 \cos (2 x) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 1.3.4.1

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

$lim_{x \to 0} \frac{12 \cos (2 x) - 4 \frac{d}{d x} [x]}{\frac{d}{d x} [x]}$

Step 1.3.4.2

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

$lim_{x \to 0} \frac{12 \cos (2 x) - 4 \cdot 1}{\frac{d}{d x} [x]}$

Step 1.3.4.3

Multiply $- 4$ by $1$ .

$lim_{x \to 0} \frac{12 \cos (2 x) - 4}{\frac{d}{d x} [x]}$

Step 1.3.5

Reorder terms.

$lim_{x \to 0} \frac{- 4 + 12 \cos (2 x)}{\frac{d}{d x} [x]}$

Step 1.3.6

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

$lim_{x \to 0} \frac{- 4 + 12 \cos (2 x)}{1}$

Step 1.4

Divide $- 4 + 12 \cos (2 x)$ by $1$ .

$lim_{x \to 0} - 4 + 12 \cos (2 x)$

Step 2

Evaluate the limit.

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Step 2.1

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

$- lim_{x \to 0} 4 + lim_{x \to 0} 12 \cos (2 x)$

Step 2.2

Evaluate the limit of $4$ which is constant as $x$ approaches $0$ .

$- 1 \cdot 4 + lim_{x \to 0} 12 \cos (2 x)$

Step 2.3

Move the term $12$ outside of the limit because it is constant with respect to $x$ .

$- 4 + 12 lim_{x \to 0} \cos (2 x)$

Step 2.4

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

$- 4 + 12 \cos (lim_{x \to 0} 2 x)$

Step 2.5

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

$- 4 + 12 \cos (2 lim_{x \to 0} x)$

Step 3

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

$- 4 + 12 \cos (2 \cdot 0)$

Step 4

Simplify the answer.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $2$ by $0$ .

$- 4 + 12 \cos (0)$

Step 4.1.2

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

$- 4 + 12 \cdot 1$

Step 4.1.3

Multiply $12$ by $1$ .

$- 4 + 12$

Step 4.2

Add $- 4$ and $12$ .

$8$