Calculus Examples

$lim_{θ \to \frac{π}{2}} \frac{16 θ - 8 π}{\cos (2 π - θ)}$

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.

$\frac{lim_{θ \to \frac{π}{2}} 16 θ - 8 π}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Evaluate the limit.

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Step 1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $θ$ approaches $\frac{π}{2}$ .

$\frac{lim_{θ \to \frac{π}{2}} 16 θ - lim_{θ \to \frac{π}{2}} 8 π}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.2.1.2

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

$\frac{16 lim_{θ \to \frac{π}{2}} θ - lim_{θ \to \frac{π}{2}} 8 π}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.2.1.3

Evaluate the limit of $8 π$ which is constant as $θ$ approaches $\frac{π}{2}$ .

$\frac{16 lim_{θ \to \frac{π}{2}} θ - 8 π}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.2.2

Evaluate the limit of $θ$ by plugging in $\frac{π}{2}$ for $θ$ .

$\frac{16 \frac{π}{2} - 8 π}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Cancel the common factor of $2$ .

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Step 1.2.3.1.1

Factor $2$ out of $16$ .

$\frac{2 (8) \frac{π}{2} - 8 π}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.2.3.1.2

Cancel the common factor.

$\frac{2 \cdot 8 \frac{π}{2} - 8 π}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.2.3.1.3

Rewrite the expression.

$\frac{8 π - 8 π}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.2.3.2

Subtract $8 π$ from $8 π$ .

$\frac{0}{lim_{θ \to \frac{π}{2}} \cos (2 π - θ)}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

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

$\frac{0}{\cos (lim_{θ \to \frac{π}{2}} 2 π - θ)}$

Step 1.3.1.2

Split the limit using the Sum of Limits Rule on the limit as $θ$ approaches $\frac{π}{2}$ .

$\frac{0}{\cos (lim_{θ \to \frac{π}{2}} 2 π - lim_{θ \to \frac{π}{2}} θ)}$

Step 1.3.1.3

Evaluate the limit of $2 π$ which is constant as $θ$ approaches $\frac{π}{2}$ .

$\frac{0}{\cos (2 π - lim_{θ \to \frac{π}{2}} θ)}$

Step 1.3.2

Evaluate the limit of $θ$ by plugging in $\frac{π}{2}$ for $θ$ .

$\frac{0}{\cos (2 π - \frac{π}{2})}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.3.3.2

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 1.3.3.3

Combine the numerators over the common denominator.

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

Step 1.3.3.4

Simplify the numerator.

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Step 1.3.3.4.1

Multiply $2$ by $2$ .

$\frac{0}{\cos (\frac{4 π - π}{2})}$

Step 1.3.3.4.2

Subtract $π$ from $4 π$ .

$\frac{0}{\cos (\frac{3 π}{2})}$

Step 1.3.3.5

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$\frac{0}{\cos (\frac{π}{2})}$

Step 1.3.3.6

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$\frac{0}{0}$

Step 1.3.3.7

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

Undefined

$\frac{0}{0}$

Step 1.3.4

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.

$lim_{θ \to \frac{π}{2}} \frac{16 θ - 8 π}{\cos (2 π - θ)} = lim_{θ \to \frac{π}{2}} \frac{\frac{d}{d θ} [16 θ - 8 π]}{\frac{d}{d θ} [\cos (2 π - θ)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{θ \to \frac{π}{2}} \frac{\frac{d}{d θ} [16 θ - 8 π]}{\frac{d}{d θ} [\cos (2 π - θ)]}$

Step 3.2

By the Sum Rule, the derivative of $16 θ - 8 π$ with respect to $θ$ is $\frac{d}{d θ} [16 θ] + \frac{d}{d θ} [- 8 π]$ .

$lim_{θ \to \frac{π}{2}} \frac{\frac{d}{d θ} [16 θ] + \frac{d}{d θ} [- 8 π]}{\frac{d}{d θ} [\cos (2 π - θ)]}$

Step 3.3

Evaluate $\frac{d}{d θ} [16 θ]$ .

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Step 3.3.1

Since $16$ is constant with respect to $θ$ , the derivative of $16 θ$ with respect to $θ$ is $16 \frac{d}{d θ} [θ]$ .

$lim_{θ \to \frac{π}{2}} \frac{16 \frac{d}{d θ} [θ] + \frac{d}{d θ} [- 8 π]}{\frac{d}{d θ} [\cos (2 π - θ)]}$

Step 3.3.2

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

$lim_{θ \to \frac{π}{2}} \frac{16 \cdot 1 + \frac{d}{d θ} [- 8 π]}{\frac{d}{d θ} [\cos (2 π - θ)]}$

Step 3.3.3

Multiply $16$ by $1$ .

$lim_{θ \to \frac{π}{2}} \frac{16 + \frac{d}{d θ} [- 8 π]}{\frac{d}{d θ} [\cos (2 π - θ)]}$

Step 3.4

Since $- 8 π$ is constant with respect to $θ$ , the derivative of $- 8 π$ with respect to $θ$ is $0$ .

$lim_{θ \to \frac{π}{2}} \frac{16 + 0}{\frac{d}{d θ} [\cos (2 π - θ)]}$

Step 3.5

Add $16$ and $0$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{\frac{d}{d θ} [\cos (2 π - θ)]}$

Step 3.6

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

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Step 3.6.1

To apply the Chain Rule, set $u$ as $2 π - θ$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{\frac{d}{d u} [\cos (u)] \frac{d}{d θ} [2 π - θ]}$

Step 3.6.2

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

$lim_{θ \to \frac{π}{2}} \frac{16}{- \sin (u) \frac{d}{d θ} [2 π - θ]}$

Step 3.6.3

Replace all occurrences of $u$ with $2 π - θ$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{- \sin (2 π - θ) \frac{d}{d θ} [2 π - θ]}$

Step 3.7

By the Sum Rule, the derivative of $2 π - θ$ with respect to $θ$ is $\frac{d}{d θ} [2 π] + \frac{d}{d θ} [- θ]$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{- \sin (2 π - θ) (\frac{d}{d θ} [2 π] + \frac{d}{d θ} [- θ])}$

Step 3.8

Since $2 π$ is constant with respect to $θ$ , the derivative of $2 π$ with respect to $θ$ is $0$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{- \sin (2 π - θ) (0 + \frac{d}{d θ} [- θ])}$

Step 3.9

Add $0$ and $\frac{d}{d θ} [- θ]$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{- \sin (2 π - θ) \frac{d}{d θ} [- θ]}$

Step 3.10

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

$lim_{θ \to \frac{π}{2}} \frac{16}{- \sin (2 π - θ) (- \frac{d}{d θ} [θ])}$

Step 3.11

Multiply $- 1$ by $- 1$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{1 \sin (2 π - θ) \frac{d}{d θ} [θ]}$

Step 3.12

Multiply $\sin (2 π - θ)$ by $1$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{\sin (2 π - θ) \frac{d}{d θ} [θ]}$

Step 3.13

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

$lim_{θ \to \frac{π}{2}} \frac{16}{\sin (2 π - θ) \cdot 1}$

Step 3.14

Multiply $\sin (2 π - θ)$ by $1$ .

$lim_{θ \to \frac{π}{2}} \frac{16}{\sin (2 π - θ)}$

Step 4

Evaluate the limit.

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Step 4.1

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

$16 lim_{θ \to \frac{π}{2}} \frac{1}{\sin (2 π - θ)}$

Step 4.2

Split the limit using the Limits Quotient Rule on the limit as $θ$ approaches $\frac{π}{2}$ .

$16 \frac{lim_{θ \to \frac{π}{2}} 1}{lim_{θ \to \frac{π}{2}} \sin (2 π - θ)}$

Step 4.3

Evaluate the limit of $1$ which is constant as $θ$ approaches $\frac{π}{2}$ .

$16 \frac{1}{lim_{θ \to \frac{π}{2}} \sin (2 π - θ)}$

Step 4.4

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

$16 \frac{1}{\sin (lim_{θ \to \frac{π}{2}} 2 π - θ)}$

Step 4.5

Split the limit using the Sum of Limits Rule on the limit as $θ$ approaches $\frac{π}{2}$ .

$16 \frac{1}{\sin (lim_{θ \to \frac{π}{2}} 2 π - lim_{θ \to \frac{π}{2}} θ)}$

Step 4.6

Evaluate the limit of $2 π$ which is constant as $θ$ approaches $\frac{π}{2}$ .

$16 \frac{1}{\sin (2 π - lim_{θ \to \frac{π}{2}} θ)}$

Step 5

Evaluate the limit of $θ$ by plugging in $\frac{π}{2}$ for $θ$ .

$16 \frac{1}{\sin (2 π - \frac{π}{2})}$

Step 6

Simplify the answer.

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Step 6.1

Convert from $\frac{1}{\sin (2 π - \frac{π}{2})}$ to $\csc (2 π - \frac{π}{2})$ .

$16 \csc (2 π - \frac{π}{2})$

Step 6.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$16 \csc (2 π \cdot \frac{2}{2} - \frac{π}{2})$

Step 6.3

Combine $2 π$ and $\frac{2}{2}$ .

$16 \csc (\frac{2 π \cdot 2}{2} - \frac{π}{2})$

Step 6.4

Combine the numerators over the common denominator.

$16 \csc (\frac{2 π \cdot 2 - π}{2})$

Step 6.5

Simplify the numerator.

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Step 6.5.1

Multiply $2$ by $2$ .

$16 \csc (\frac{4 π - π}{2})$

Step 6.5.2

Subtract $π$ from $4 π$ .

$16 \csc (\frac{3 π}{2})$

Step 6.6

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosecant is negative in the fourth quadrant.

$16 (- \csc (\frac{π}{2}))$

Step 6.7

The exact value of $\csc (\frac{π}{2})$ is $1$ .

$16 (- 1 \cdot 1)$

Step 6.8

Multiply $16 (- 1 \cdot 1)$ .

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Step 6.8.1

Multiply $- 1$ by $1$ .

$16 \cdot - 1$

Step 6.8.2

Multiply $16$ by $- 1$ .

$- 16$