Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

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

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

Step 1.1.2.1.2

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

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

Step 1.1.2.1.3

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

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

Step 1.1.2.1.4

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

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

Step 1.1.2.1.5

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Apply pythagorean identity.

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

Step 1.1.2.3.2

Multiply $2$ by $0$ .

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

Step 1.1.2.3.3

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

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

Step 1.1.2.3.4

Raising $0$ to any positive power yields $0$ .

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

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

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

Step 1.1.3.1.2

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

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

Step 1.1.3.2

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

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

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Multiply $2$ by $0$ .

$\frac{0}{0^{2}}$

Step 1.1.3.3.2

Raising $0$ to any positive power yields $0$ .

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

By the Sum Rule, the derivative of $1 - \cos^{2} (2 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \cos^{2} (2 x)]$ .

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

Step 1.3.3

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

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

Step 1.3.4

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

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Step 1.3.4.1

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

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

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 $f (x) = x^{2}$ and $g (x) = \cos (2 x)$ .

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Step 1.3.4.2.1

To apply the Chain Rule, set $u_{1}$ as $\cos (2 x)$ .

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

Step 1.3.4.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

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

Step 1.3.4.2.3

Replace all occurrences of $u_{1}$ with $\cos (2 x)$ .

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

Step 1.3.4.3

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 $g (x) = 2 x$ .

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Step 1.3.4.3.1

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

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

Step 1.3.4.3.2

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

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

Step 1.3.4.3.3

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

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

Step 1.3.4.4

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

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

Step 1.3.4.5

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

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

Step 1.3.4.6

Multiply $2$ by $1$ .

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

Step 1.3.4.7

Multiply $2$ by $- 1$ .

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

Step 1.3.4.8

Multiply $- 2$ by $2$ .

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

Step 1.3.4.9

Multiply $- 4$ by $- 1$ .

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

Step 1.3.5

Add $0$ and $4 \cos (2 x) \sin (2 x)$ .

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

Step 1.3.6

Apply the product rule to $2 x$ .

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

Step 1.3.7

Raise $2$ to the power of $2$ .

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

Step 1.3.8

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

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

Step 1.3.9

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 1.3.10

Multiply $2$ by $4$ .

$lim_{x \to 0} \frac{4 \cos (2 x) \sin (2 x)}{8 x}$

Step 1.4

Cancel the common factor of $4$ and $8$ .

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Step 1.4.1

Factor $4$ out of $4 \cos (2 x) \sin (2 x)$ .

$lim_{x \to 0} \frac{4 (\cos (2 x) \sin (2 x))}{8 x}$

Step 1.4.2

Cancel the common factors.

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Step 1.4.2.1

Factor $4$ out of $8 x$ .

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

Step 1.4.2.2

Cancel the common factor.

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

Step 1.4.2.3

Rewrite the expression.

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

Step 2

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

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

Step 3

Apply L'Hospital's rule.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

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

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

Step 3.1.2.2

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

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

Step 3.1.2.3

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

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

Step 3.1.2.4

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

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

Step 3.1.2.5

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

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

Step 3.1.2.6

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

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Step 3.1.2.6.1

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

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

Step 3.1.2.6.2

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

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

Step 3.1.2.7

Simplify the answer.

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Step 3.1.2.7.1

Multiply $2$ by $0$ .

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

Step 3.1.2.7.2

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

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

Step 3.1.2.7.3

Multiply $\sin (2 \cdot 0)$ by $1$ .

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

Step 3.1.2.7.4

Multiply $2$ by $0$ .

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

Step 3.1.2.7.5

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

$\frac{1}{2} \cdot \frac{0}{lim_{x \to 0} x}$

Step 3.1.3

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

$\frac{1}{2} \cdot \frac{0}{0}$

Step 3.1.4

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

Undefined

$\frac{1}{2} \cdot \frac{0}{0}$

Step 3.2

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.

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

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

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

Step 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 $f (x) = \cos (2 x)$ and $g (x) = \sin (2 x)$ .

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

Step 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 $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 3.3.3.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

Step 3.3.3.2

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

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

Step 3.3.3.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 3.3.4

Raise $\cos (2 x)$ to the power of $1$ .

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

Step 3.3.5

Raise $\cos (2 x)$ to the power of $1$ .

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

Step 3.3.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.3.7

Add $1$ and $1$ .

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

Step 3.3.8

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

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

Step 3.3.9

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

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

Step 3.3.10

Multiply $2$ by $1$ .

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

Step 3.3.11

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

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

Step 3.3.12

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 3.3.12.1

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

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

Step 3.3.12.2

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

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

Step 3.3.12.3

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

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

Step 3.3.13

Raise $\sin (2 x)$ to the power of $1$ .

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

Step 3.3.14

Raise $\sin (2 x)$ to the power of $1$ .

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

Step 3.3.15

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.3.16

Add $1$ and $1$ .

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

Step 3.3.17

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

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

Step 3.3.18

Multiply $2$ by $- 1$ .

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

Step 3.3.19

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

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

Step 3.3.20

Multiply $- 2$ by $1$ .

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

Step 3.3.21

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

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

Step 3.4

Divide $2 \cos^{2} (2 x) - 2 \sin^{2} (2 x)$ by $1$ .

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

Step 4

Evaluate the limit.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

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

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

Step 4.5

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

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

Step 4.6

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

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

Step 4.7

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

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

Step 4.8

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

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

Step 4.9

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

$\frac{1}{2} (2 \cos^{2} (2 \cdot 0) - 2 \sin^{2} (2 \cdot 0))$

Step 6

Simplify the answer.

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Step 6.1

Simplify each term.

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Step 6.1.1

Multiply $2$ by $0$ .

$\frac{1}{2} (2 \cos^{2} (0) - 2 \sin^{2} (2 \cdot 0))$

Step 6.1.2

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

$\frac{1}{2} (2 \cdot 1^{2} - 2 \sin^{2} (2 \cdot 0))$

Step 6.1.3

One to any power is one.

$\frac{1}{2} (2 \cdot 1 - 2 \sin^{2} (2 \cdot 0))$

Step 6.1.4

Multiply $2$ by $1$ .

$\frac{1}{2} (2 - 2 \sin^{2} (2 \cdot 0))$

Step 6.1.5

Multiply $2$ by $0$ .

$\frac{1}{2} (2 - 2 \sin^{2} (0))$

Step 6.1.6

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

$\frac{1}{2} (2 - 2 \cdot 0^{2})$

Step 6.1.7

Raising $0$ to any positive power yields $0$ .

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

Step 6.1.8

Multiply $- 2$ by $0$ .

$\frac{1}{2} (2 + 0)$

Step 6.2

Add $2$ and $0$ .

$\frac{1}{2} \cdot 2$

Step 6.3

Cancel the common factor of $2$ .

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Step 6.3.1

Cancel the common factor.

$\frac{1}{2} \cdot 2$

Step 6.3.2

Rewrite the expression.

$1$