Calculus Examples

$lim_{x \to 0} \frac{2 x \cos (8 x) + 2 \sin (5 x)}{\tan (2 x)}$

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.

$\frac{lim_{x \to 0} 2 x \cos (8 x) + 2 \sin (5 x)}{lim_{x \to 0} \tan (2 x)}$

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 $0$ .

$\frac{lim_{x \to 0} 2 x \cos (8 x) + lim_{x \to 0} 2 \sin (5 x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.2

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

$\frac{2 lim_{x \to 0} x \cos (8 x) + lim_{x \to 0} 2 \sin (5 x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.3

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

$\frac{2 lim_{x \to 0} x \cdot lim_{x \to 0} \cos (8 x) + lim_{x \to 0} 2 \sin (5 x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.4

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

$\frac{2 lim_{x \to 0} x \cdot \cos (lim_{x \to 0} 8 x) + lim_{x \to 0} 2 \sin (5 x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.5

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

$\frac{2 lim_{x \to 0} x \cdot \cos (8 lim_{x \to 0} x) + lim_{x \to 0} 2 \sin (5 x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.6

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

$\frac{2 lim_{x \to 0} x \cdot \cos (8 lim_{x \to 0} x) + 2 lim_{x \to 0} \sin (5 x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.7

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

$\frac{2 lim_{x \to 0} x \cdot \cos (8 lim_{x \to 0} x) + 2 \sin (lim_{x \to 0} 5 x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.8

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{2 lim_{x \to 0} x \cdot \cos (8 lim_{x \to 0} x) + 2 \sin (5 lim_{x \to 0} x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.9

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

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Step 1.1.2.9.1

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

$\frac{2 \cdot 0 \cdot \cos (8 lim_{x \to 0} x) + 2 \sin (5 lim_{x \to 0} x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.9.2

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

$\frac{2 \cdot 0 \cdot \cos (8 \cdot 0) + 2 \sin (5 lim_{x \to 0} x)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.9.3

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

$\frac{2 \cdot 0 \cdot \cos (8 \cdot 0) + 2 \sin (5 \cdot 0)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.10

Simplify the answer.

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Step 1.1.2.10.1

Simplify each term.

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Step 1.1.2.10.1.1

Multiply $2$ by $0$ .

$\frac{0 \cdot \cos (8 \cdot 0) + 2 \sin (5 \cdot 0)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.10.1.2

Multiply $8$ by $0$ .

$\frac{0 \cdot \cos (0) + 2 \sin (5 \cdot 0)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.10.1.3

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

$\frac{0 \cdot 1 + 2 \sin (5 \cdot 0)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.10.1.4

Multiply $0$ by $1$ .

$\frac{0 + 2 \sin (5 \cdot 0)}{lim_{x \to 0} \tan (2 x)}$

Step 1.1.2.10.1.5

Multiply $5$ by $0$ .

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

Step 1.1.2.10.1.6

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

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

Step 1.1.2.10.1.7

Multiply $2$ by $0$ .

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

Step 1.1.2.10.2

Add $0$ and $0$ .

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

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

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

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

Step 1.1.3.1.2

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

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

Step 1.1.3.2

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

$\frac{0}{\tan (2 \cdot 0)}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Multiply $2$ by $0$ .

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

Step 1.1.3.3.2

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

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.4

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{2 x \cos (8 x) + 2 \sin (5 x)}{\tan (2 x)} = lim_{x \to 0} \frac{\frac{d}{d x} [2 x \cos (8 x) + 2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [2 x \cos (8 x) + 2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.2

By the Sum Rule, the derivative of $2 x \cos (8 x) + 2 \sin (5 x)$ with respect to $x$ is $\frac{d}{d x} [2 x \cos (8 x)] + \frac{d}{d x} [2 \sin (5 x)]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [2 x \cos (8 x)] + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [2 x \cos (8 x)]$ .

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Step 1.3.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x \cos (8 x)$ with respect to $x$ is $2 \frac{d}{d x} [x \cos (8 x)]$ .

$lim_{x \to 0} \frac{2 \frac{d}{d x} [x \cos (8 x)] + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \cos (8 x)$ .

$lim_{x \to 0} \frac{2 (x \frac{d}{d x} [\cos (8 x)] + \cos (8 x) \frac{d}{d x} [x]) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.3

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 $g (x) = 8 x$ .

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Step 1.3.3.3.1

To apply the Chain Rule, set $u_{1}$ as $8 x$ .

$lim_{x \to 0} \frac{2 (x (\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d x} [8 x]) + \cos (8 x) \frac{d}{d x} [x]) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.3.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$lim_{x \to 0} \frac{2 (x (- \sin (u_{1}) \frac{d}{d x} [8 x]) + \cos (8 x) \frac{d}{d x} [x]) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.3.3

Replace all occurrences of $u_{1}$ with $8 x$ .

$lim_{x \to 0} \frac{2 (x (- \sin (8 x) \frac{d}{d x} [8 x]) + \cos (8 x) \frac{d}{d x} [x]) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.4

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{2 (x (- \sin (8 x) (8 \frac{d}{d x} [x])) + \cos (8 x) \frac{d}{d x} [x]) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.5

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

$lim_{x \to 0} \frac{2 (x (- \sin (8 x) (8 \cdot 1)) + \cos (8 x) \frac{d}{d x} [x]) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.6

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

$lim_{x \to 0} \frac{2 (x (- \sin (8 x) (8 \cdot 1)) + \cos (8 x) \cdot 1) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.7

Multiply $8$ by $1$ .

$lim_{x \to 0} \frac{2 (x (- \sin (8 x) \cdot 8) + \cos (8 x) \cdot 1) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.8

Multiply $8$ by $- 1$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x) \cdot 1) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.3.9

Multiply $\cos (8 x)$ by $1$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + \frac{d}{d x} [2 \sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [2 \sin (5 x)]$ .

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Step 1.3.4.1

Since $2$ is constant with respect to $x$ , the derivative of $2 \sin (5 x)$ with respect to $x$ is $2 \frac{d}{d x} [\sin (5 x)]$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 2 \frac{d}{d x} [\sin (5 x)]}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4.2

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 $g (x) = 5 x$ .

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Step 1.3.4.2.1

To apply the Chain Rule, set $u_{2}$ as $5 x$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 2 (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d x} [5 x])}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4.2.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 2 (\cos (u_{2}) \frac{d}{d x} [5 x])}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4.2.3

Replace all occurrences of $u_{2}$ with $5 x$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 2 (\cos (5 x) \frac{d}{d x} [5 x])}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4.3

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 2 (\cos (5 x) (5 \frac{d}{d x} [x]))}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4.4

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

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 2 (\cos (5 x) (5 \cdot 1))}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4.5

Multiply $5$ by $1$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 2 (\cos (5 x) \cdot 5)}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4.6

Move $5$ to the left of $\cos (5 x)$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 2 (5 \cdot \cos (5 x))}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.4.7

Multiply $5$ by $2$ .

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x)) + \cos (8 x)) + 10 \cos (5 x)}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.5

Simplify.

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Step 1.3.5.1

Apply the distributive property.

$lim_{x \to 0} \frac{2 (x (- 8 \sin (8 x))) + 2 \cos (8 x) + 10 \cos (5 x)}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.5.2

Multiply $- 8$ by $2$ .

$lim_{x \to 0} \frac{- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)}{\frac{d}{d x} [\tan (2 x)]}$

Step 1.3.6

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

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Step 1.3.6.1

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

$lim_{x \to 0} \frac{- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)}{\frac{d}{d u} [\tan (u)] \frac{d}{d x} [2 x]}$

Step 1.3.6.2

The derivative of $\tan (u)$ with respect to $u$ is $\sec^{2} (u)$ .

$lim_{x \to 0} \frac{- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)}{\sec^{2} (u) \frac{d}{d x} [2 x]}$

Step 1.3.6.3

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

$lim_{x \to 0} \frac{- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)}{\sec^{2} (2 x) \frac{d}{d x} [2 x]}$

Step 1.3.7

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

$lim_{x \to 0} \frac{- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)}{\sec^{2} (2 x) (2 \frac{d}{d x} [x])}$

Step 1.3.8

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

$lim_{x \to 0} \frac{- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)}{\sec^{2} (2 x) (2 \cdot 1)}$

Step 1.3.9

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)}{\sec^{2} (2 x) \cdot 2}$

Step 1.3.10

Move $2$ to the left of $\sec^{2} (2 x)$ .

$lim_{x \to 0} \frac{- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)}{2 \sec^{2} (2 x)}$

Step 1.4

Cancel the common factor of $- 16 x \sin (8 x) + 2 \cos (8 x) + 10 \cos (5 x)$ and $2$ .

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Step 1.4.1

Factor $2$ out of $- 16 x \sin (8 x)$ .

$lim_{x \to 0} \frac{2 (- 8 x \sin (8 x)) + 2 \cos (8 x) + 10 \cos (5 x)}{2 \sec^{2} (2 x)}$

Step 1.4.2

Factor $2$ out of $2 \cos (8 x)$ .

$lim_{x \to 0} \frac{2 (- 8 x \sin (8 x)) + 2 (\cos (8 x)) + 10 \cos (5 x)}{2 \sec^{2} (2 x)}$

Step 1.4.3

Factor $2$ out of $2 (- 8 x \sin (8 x)) + 2 (\cos (8 x))$ .

$lim_{x \to 0} \frac{2 (- 8 x \sin (8 x) + \cos (8 x)) + 10 \cos (5 x)}{2 \sec^{2} (2 x)}$

Step 1.4.4

Factor $2$ out of $10 \cos (5 x)$ .

$lim_{x \to 0} \frac{2 (- 8 x \sin (8 x) + \cos (8 x)) + 2 (5 \cos (5 x))}{2 \sec^{2} (2 x)}$

Step 1.4.5

Factor $2$ out of $2 (- 8 x \sin (8 x) + \cos (8 x)) + 2 (5 \cos (5 x))$ .

$lim_{x \to 0} \frac{2 (- 8 x \sin (8 x) + \cos (8 x) + 5 \cos (5 x))}{2 \sec^{2} (2 x)}$

Step 1.4.6

Cancel the common factors.

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Step 1.4.6.1

Factor $2$ out of $2 \sec^{2} (2 x)$ .

$lim_{x \to 0} \frac{2 (- 8 x \sin (8 x) + \cos (8 x) + 5 \cos (5 x))}{2 (\sec^{2} (2 x))}$

Step 1.4.6.2

Cancel the common factor.

$lim_{x \to 0} \frac{2 (- 8 x \sin (8 x) + \cos (8 x) + 5 \cos (5 x))}{2 \sec^{2} (2 x)}$

Step 1.4.6.3

Rewrite the expression.

$lim_{x \to 0} \frac{- 8 x \sin (8 x) + \cos (8 x) + 5 \cos (5 x)}{\sec^{2} (2 x)}$

Step 2

Evaluate the limit.

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Step 2.1

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

$\frac{lim_{x \to 0} - 8 x \sin (8 x) + \cos (8 x) + 5 \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.2

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

$\frac{- lim_{x \to 0} 8 x \sin (8 x) + lim_{x \to 0} \cos (8 x) + lim_{x \to 0} 5 \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.3

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

$\frac{- 8 lim_{x \to 0} x \sin (8 x) + lim_{x \to 0} \cos (8 x) + lim_{x \to 0} 5 \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.4

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

$\frac{- 8 lim_{x \to 0} x \cdot lim_{x \to 0} \sin (8 x) + lim_{x \to 0} \cos (8 x) + lim_{x \to 0} 5 \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.5

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

$\frac{- 8 lim_{x \to 0} x \cdot \sin (lim_{x \to 0} 8 x) + lim_{x \to 0} \cos (8 x) + lim_{x \to 0} 5 \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.6

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

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + lim_{x \to 0} \cos (8 x) + lim_{x \to 0} 5 \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.7

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

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + \cos (lim_{x \to 0} 8 x) + lim_{x \to 0} 5 \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.8

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

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + \cos (8 lim_{x \to 0} x) + lim_{x \to 0} 5 \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.9

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + \cos (8 lim_{x \to 0} x) + 5 lim_{x \to 0} \cos (5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.10

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

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + \cos (8 lim_{x \to 0} x) + 5 \cos (lim_{x \to 0} 5 x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.11

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + \cos (8 lim_{x \to 0} x) + 5 \cos (5 lim_{x \to 0} x)}{lim_{x \to 0} \sec^{2} (2 x)}$

Step 2.12

Move the exponent $2$ from $\sec^{2} (2 x)$ outside the limit using the Limits Power Rule.

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + \cos (8 lim_{x \to 0} x) + 5 \cos (5 lim_{x \to 0} x)}{{(lim_{x \to 0} \sec (2 x))}^{2}}$

Step 2.13

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

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + \cos (8 lim_{x \to 0} x) + 5 \cos (5 lim_{x \to 0} x)}{\sec^{2} (lim_{x \to 0} 2 x)}$

Step 2.14

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

$\frac{- 8 lim_{x \to 0} x \cdot \sin (8 lim_{x \to 0} x) + \cos (8 lim_{x \to 0} x) + 5 \cos (5 lim_{x \to 0} x)}{\sec^{2} (2 lim_{x \to 0} x)}$

Step 3

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

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Step 3.1

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

$\frac{- 8 \cdot 0 \cdot \sin (8 lim_{x \to 0} x) + \cos (8 lim_{x \to 0} x) + 5 \cos (5 lim_{x \to 0} x)}{\sec^{2} (2 lim_{x \to 0} x)}$

Step 3.2

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

$\frac{- 8 \cdot 0 \cdot \sin (8 \cdot 0) + \cos (8 lim_{x \to 0} x) + 5 \cos (5 lim_{x \to 0} x)}{\sec^{2} (2 lim_{x \to 0} x)}$

Step 3.3

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

$\frac{- 8 \cdot 0 \cdot \sin (8 \cdot 0) + \cos (8 \cdot 0) + 5 \cos (5 lim_{x \to 0} x)}{\sec^{2} (2 lim_{x \to 0} x)}$

Step 3.4

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

$\frac{- 8 \cdot 0 \cdot \sin (8 \cdot 0) + \cos (8 \cdot 0) + 5 \cos (5 \cdot 0)}{\sec^{2} (2 lim_{x \to 0} x)}$

Step 3.5

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

$\frac{- 8 \cdot 0 \cdot \sin (8 \cdot 0) + \cos (8 \cdot 0) + 5 \cos (5 \cdot 0)}{\sec^{2} (2 \cdot 0)}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $- 8$ by $0$ .

$\frac{0 \cdot \sin (8 \cdot 0) + \cos (8 \cdot 0) + 5 \cos (5 \cdot 0)}{\sec^{2} (2 \cdot 0)}$

Step 4.1.2

Multiply $8$ by $0$ .

$\frac{0 \cdot \sin (0) + \cos (8 \cdot 0) + 5 \cos (5 \cdot 0)}{\sec^{2} (2 \cdot 0)}$

Step 4.1.3

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

$\frac{0 \cdot 0 + \cos (8 \cdot 0) + 5 \cos (5 \cdot 0)}{\sec^{2} (2 \cdot 0)}$

Step 4.1.4

Multiply $0$ by $0$ .

$\frac{0 + \cos (8 \cdot 0) + 5 \cos (5 \cdot 0)}{\sec^{2} (2 \cdot 0)}$

Step 4.1.5

Multiply $8$ by $0$ .

$\frac{0 + \cos (0) + 5 \cos (5 \cdot 0)}{\sec^{2} (2 \cdot 0)}$

Step 4.1.6

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

$\frac{0 + 1 + 5 \cos (5 \cdot 0)}{\sec^{2} (2 \cdot 0)}$

Step 4.1.7

Multiply $5$ by $0$ .

$\frac{0 + 1 + 5 \cos (0)}{\sec^{2} (2 \cdot 0)}$

Step 4.1.8

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

$\frac{0 + 1 + 5 \cdot 1}{\sec^{2} (2 \cdot 0)}$

Step 4.1.9

Multiply $5$ by $1$ .

$\frac{0 + 1 + 5}{\sec^{2} (2 \cdot 0)}$

Step 4.1.10

Add $0$ and $1$ .

$\frac{1 + 5}{\sec^{2} (2 \cdot 0)}$

Step 4.1.11

Add $1$ and $5$ .

$\frac{6}{\sec^{2} (2 \cdot 0)}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $0$ .

$\frac{6}{\sec^{2} (0)}$

Step 4.2.2

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

$\frac{6}{1^{2}}$

Step 4.2.3

One to any power is one.

$\frac{6}{1}$

Step 4.3

Divide $6$ by $1$ .

$6$