Evaluate Using L’Hospital’s Rule limit as x approaches 0 of (1-cos(4x))/(9x^2)

$lim_{x \to 0} \frac{(1 - \cos (4 x))}{9 x^{2}}$

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

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

$\frac{lim_{x \to 0} 1 - \cos (4 x)}{lim_{x \to 0} 9 x^{2}}$

Evaluate the limit of the numerator.

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Take the limit of each term.

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

$\frac{lim_{x \to 0} 1 - lim_{x \to 0} \cos (4 x)}{lim_{x \to 0} 9 x^{2}}$

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

$\frac{lim_{x \to 0} 1 - \cos (lim_{x \to 0} 4 x)}{lim_{x \to 0} 9 x^{2}}$

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

$\frac{lim_{x \to 0} 1 - \cos (4 lim_{x \to 0} x)}{lim_{x \to 0} 9 x^{2}}$

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

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

$\frac{1 - \cos (4 lim_{x \to 0} x)}{lim_{x \to 0} 9 x^{2}}$

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

$\frac{1 - \cos (4 \cdot 0)}{lim_{x \to 0} 9 x^{2}}$

Simplify the answer.

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Simplify each term.

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Multiply $4$ by $0$ .

$\frac{1 - \cos (0)}{lim_{x \to 0} 9 x^{2}}$

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

$\frac{1 - 1 \cdot 1}{lim_{x \to 0} 9 x^{2}}$

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to 0} 9 x^{2}}$

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 0} 9 x^{2}}$

Evaluate the limit of the denominator.

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Take the limit of each term.

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Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{9 lim_{x \to 0} x^{2}}$

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

$\frac{0}{9 {(lim_{x \to 0} x)}^{2}}$

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

$\frac{0}{9 \cdot 0^{2}}$

Simplify the answer.

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

$\frac{0}{9 \cdot 0}$

Multiply $9$ by $0$ .

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{1 - \cos (4 x)}{9 x^{2}} = lim_{x \to 0} \frac{\frac{d}{d x} [1 - \cos (4 x)]}{\frac{d}{d x} [9 x^{2}]}$

Find the derivative of the numerator and denominator.

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Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [1 - \cos (4 x)]}{\frac{d}{d x} [9 x^{2}]}$

By the Sum Rule, the derivative of $1 - \cos (4 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (4 x)]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (4 x)]}{\frac{d}{d x} [9 x^{2}]}$

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

$lim_{x \to 0} \frac{0 + \frac{d}{d x} [- \cos (4 x)]}{\frac{d}{d x} [9 x^{2}]}$

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

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Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (4 x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (4 x)]$ .

$lim_{x \to 0} \frac{0 - \frac{d}{d x} [\cos (4 x)]}{\frac{d}{d x} [9 x^{2}]}$

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 $g (x) = 4 x$ .

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To apply the Chain Rule, set $u$ as $4 x$ .

$lim_{x \to 0} \frac{0 - (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [4 x])}{\frac{d}{d x} [9 x^{2}]}$

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

$lim_{x \to 0} \frac{0 - (- \sin (u) \frac{d}{d x} [4 x])}{\frac{d}{d x} [9 x^{2}]}$

Replace all occurrences of $u$ with $4 x$ .

$lim_{x \to 0} \frac{0 - (- \sin (4 x) \frac{d}{d x} [4 x])}{\frac{d}{d x} [9 x^{2}]}$

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

$lim_{x \to 0} \frac{0 - (- \sin (4 x) (4 \frac{d}{d x} [x]))}{\frac{d}{d x} [9 x^{2}]}$

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

$lim_{x \to 0} \frac{0 - (- \sin (4 x) (4 \cdot 1))}{\frac{d}{d x} [9 x^{2}]}$

Multiply $4$ by $1$ .

$lim_{x \to 0} \frac{0 - (- \sin (4 x) \cdot 4)}{\frac{d}{d x} [9 x^{2}]}$

Multiply $4$ by $- 1$ .

$lim_{x \to 0} \frac{0 - (- 4 \sin (4 x))}{\frac{d}{d x} [9 x^{2}]}$

Multiply $- 4$ by $- 1$ .

$lim_{x \to 0} \frac{0 + 4 \sin (4 x)}{\frac{d}{d x} [9 x^{2}]}$

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

$lim_{x \to 0} \frac{4 \sin (4 x)}{\frac{d}{d x} [9 x^{2}]}$

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to 0} \frac{4 \sin (4 x)}{9 \frac{d}{d x} [x^{2}]}$

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

$lim_{x \to 0} \frac{4 \sin (4 x)}{9 (2 x)}$

Multiply $2$ by $9$ .

$lim_{x \to 0} \frac{4 \sin (4 x)}{18 x}$

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

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

$\frac{lim_{x \to 0} 4 \sin (4 x)}{lim_{x \to 0} 18 x}$

Evaluate the limit of the numerator.

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Take the limit of each term.

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

$\frac{4 lim_{x \to 0} \sin (4 x)}{lim_{x \to 0} 18 x}$

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

$\frac{4 \sin (lim_{x \to 0} 4 x)}{lim_{x \to 0} 18 x}$

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

$\frac{4 \sin (4 lim_{x \to 0} x)}{lim_{x \to 0} 18 x}$

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

$\frac{4 \sin (4 \cdot 0)}{lim_{x \to 0} 18 x}$

Simplify the answer.

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Multiply $4$ by $0$ .

$\frac{4 \sin (0)}{lim_{x \to 0} 18 x}$

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

$\frac{4 \cdot 0}{lim_{x \to 0} 18 x}$

Multiply $4$ by $0$ .

$\frac{0}{lim_{x \to 0} 18 x}$

Evaluate the limit of the denominator.

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Move the term $18$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{18 lim_{x \to 0} x}$

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

$\frac{0}{18 \cdot 0}$

Multiply $18$ by $0$ .

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

$lim_{x \to 0} \frac{4 \sin (4 x)}{18 x} = lim_{x \to 0} \frac{\frac{d}{d x} [4 \sin (4 x)]}{\frac{d}{d x} [18 x]}$

Find the derivative of the numerator and denominator.

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Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [4 \sin (4 x)]}{\frac{d}{d x} [18 x]}$

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

$lim_{x \to 0} \frac{4 \frac{d}{d x} [\sin (4 x)]}{\frac{d}{d x} [18 x]}$

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 $g (x) = 4 x$ .

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To apply the Chain Rule, set $u$ as $4 x$ .

$lim_{x \to 0} \frac{4 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [4 x])}{\frac{d}{d x} [18 x]}$

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

$lim_{x \to 0} \frac{4 (\cos (u) \frac{d}{d x} [4 x])}{\frac{d}{d x} [18 x]}$

Replace all occurrences of $u$ with $4 x$ .

$lim_{x \to 0} \frac{4 (\cos (4 x) \frac{d}{d x} [4 x])}{\frac{d}{d x} [18 x]}$

Remove parentheses.

$lim_{x \to 0} \frac{4 \cos (4 x) \frac{d}{d x} [4 x]}{\frac{d}{d x} [18 x]}$

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

$lim_{x \to 0} \frac{4 \cos (4 x) (4 \frac{d}{d x} [x])}{\frac{d}{d x} [18 x]}$

Multiply $4$ by $4$ .

$lim_{x \to 0} \frac{16 \cos (4 x) \frac{d}{d x} [x]}{\frac{d}{d x} [18 x]}$

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

$lim_{x \to 0} \frac{16 \cos (4 x) \cdot 1}{\frac{d}{d x} [18 x]}$

Multiply $16$ by $1$ .

$lim_{x \to 0} \frac{16 \cos (4 x)}{\frac{d}{d x} [18 x]}$

Since $18$ is constant with respect to $x$ , the derivative of $18 x$ with respect to $x$ is $18 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{16 \cos (4 x)}{18 \frac{d}{d x} [x]}$

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

$lim_{x \to 0} \frac{16 \cos (4 x)}{18 \cdot 1}$

Multiply $18$ by $1$ .

$lim_{x \to 0} \frac{16 \cos (4 x)}{18}$

Take the limit of each term.

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

$\frac{(lim_{x \to 0} 16 \cos (4 x))}{(lim_{x \to 0} 18)}$

Move the term $16$ outside of the limit because it is constant with respect to $x$ .

$\frac{16 lim_{x \to 0} \cos (4 x)}{lim_{x \to 0} 18}$

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

$\frac{16 \cos (lim_{x \to 0} 4 x)}{lim_{x \to 0} 18}$

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

$\frac{16 \cos (4 lim_{x \to 0} x)}{lim_{x \to 0} 18}$

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

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

$\frac{16 \cos (4 \cdot 0)}{lim_{x \to 0} 18}$

Evaluate the limit of $18$ which is constant as $x$ approaches $0$ .

$\frac{16 \cos (4 \cdot 0)}{18}$

Simplify the answer.

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Cancel the common factor of $16$ and $18$ .

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Factor $2$ out of $16 \cos (4 \cdot 0)$ .

$\frac{2 (8 \cos (4 \cdot 0))}{18}$

Cancel the common factors.

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Factor $2$ out of $18$ .

$\frac{2 (8 \cos (4 \cdot 0))}{2 (9)}$

Cancel the common factor.

$\frac{2 (8 \cos (4 \cdot 0))}{2 \cdot 9}$

Rewrite the expression.

$\frac{8 \cos (4 \cdot 0)}{9}$

Simplify the numerator.

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Multiply $4$ by $0$ .

$\frac{8 \cos (0)}{9}$

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

$\frac{8 \cdot 1}{9}$

Multiply $8$ by $1$ .

$\frac{8}{9}$

Evaluate Using L’Hospital’s Rule limit as x approaches 0 of (1-cos(4x))/(9x^2)

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