Calculus Examples

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

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.

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

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 $θ$ approaches $0$ .

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

Step 1.1.2.1.2

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

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

Step 1.1.2.1.3

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

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

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

Step 1.1.2.3.1.2

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{θ \to 0} \sin (2 θ)}$

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{θ \to 0} \sin (2 θ)}$

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 sine is continuous.

$\frac{0}{\sin (lim_{θ \to 0} 2 θ)}$

Step 1.1.3.1.2

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

$\frac{0}{\sin (2 lim_{θ \to 0} θ)}$

Step 1.1.3.2

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

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

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Multiply $2$ by $0$ .

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

Step 1.1.3.3.2

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

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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Step 1.3.4.1

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

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

Step 1.3.4.2

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

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

Step 1.3.4.3

Multiply $- 1$ by $- 1$ .

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

Step 1.3.4.4

Multiply $\sin (θ)$ by $1$ .

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

Step 1.3.5

Add $0$ and $\sin (θ)$ .

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

Step 1.3.6

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

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Step 1.3.6.1

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

$lim_{θ \to 0} \frac{\sin (θ)}{\frac{d}{d u} [\sin (u)] \frac{d}{d θ} [2 θ]}$

Step 1.3.6.2

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

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

Step 1.3.6.3

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

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

Step 1.3.7

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

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

Step 1.3.8

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

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

Step 1.3.9

Multiply $2$ by $1$ .

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

Step 1.3.10

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

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

Step 2

Evaluate the limit.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

Simplify the answer.

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Step 4.1

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

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

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $0$ .

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

Step 4.2.2

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

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

Step 4.3

Divide $0$ by $1$ .

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

Step 4.4

Multiply $\frac{1}{2}$ by $0$ .

$0$