Calculus Examples

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

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.

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

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 $0$ .

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

Step 1.1.2.1.2

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

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

Step 1.1.2.1.3

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

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

Step 1.1.2.1.4

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

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

Step 1.1.2.1.5

Evaluate the limit of $2$ which is constant as $θ$ approaches $0$ .

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

Step 1.1.2.1.6

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

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

Step 1.1.2.1.7

Evaluate the limit of $\frac{1}{2}$ which is constant as $θ$ approaches $0$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify the denominator.

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Step 1.1.2.3.1.1

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

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

Step 1.1.2.3.1.2

Add $2$ and $0$ .

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

Step 1.1.2.3.2

Combine the numerators over the common denominator.

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

Step 1.1.2.3.3

Subtract $1$ from $1$ .

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

Step 1.1.2.3.4

Divide $0$ by $2$ .

$\frac{0}{lim_{θ \to 0} \sin (θ)}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

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

$\frac{0}{\sin (lim_{θ \to 0} θ)}$

Step 1.1.3.2

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

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

Step 1.1.3.3

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

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

Evaluate $\frac{d}{d θ} [\frac{1}{2 + \sin (θ)}]$ .

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Step 1.3.3.1

Rewrite $\frac{1}{2 + \sin (θ)}$ as ${(2 + \sin (θ))}^{- 1}$ .

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

Step 1.3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d θ} [f (g (θ))]$ is $f' (g (θ)) g' (θ)$ where $f (θ) = θ^{- 1}$ and $g (θ) = 2 + \sin (θ)$ .

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Step 1.3.3.2.1

To apply the Chain Rule, set $u$ as $2 + \sin (θ)$ .

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

Step 1.3.3.2.2

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

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

Step 1.3.3.2.3

Replace all occurrences of $u$ with $2 + \sin (θ)$ .

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

Step 1.3.3.3

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

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

Step 1.3.3.4

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

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

Step 1.3.3.5

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

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

Step 1.3.3.6

Add $0$ and $\cos (θ)$ .

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

Step 1.3.4

Since $- \frac{1}{2}$ is constant with respect to $θ$ , the derivative of $- \frac{1}{2}$ with respect to $θ$ is $0$ .

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

Step 1.3.5

Simplify.

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Step 1.3.5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.3.5.2

Combine terms.

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Step 1.3.5.2.1

Combine $\cos (θ)$ and $\frac{1}{{(2 + \sin (θ))}^{2}}$ .

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

Step 1.3.5.2.2

Add $- \frac{\cos (θ)}{{(2 + \sin (θ))}^{2}}$ and $0$ .

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

Step 1.3.6

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

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

Step 1.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5

Multiply $\frac{1}{\cos (θ)}$ by $\frac{\cos (θ)}{{(2 + \sin (θ))}^{2}}$ .

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

Step 1.6

Cancel the common factor of $\cos (θ)$ .

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Step 1.6.1

Cancel the common factor.

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

Step 1.6.2

Rewrite the expression.

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

Step 2

Evaluate the limit.

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Step 2.1

Move the term $- 1$ outside of the limit because it is constant with respect to $θ$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Evaluate the limit of $2$ which is constant as $θ$ approaches $0$ .

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

Step 2.7

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

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

Step 3

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

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

Step 4

Simplify the denominator.

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Step 4.1

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

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

Step 4.2

Add $2$ and $0$ .

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

Step 4.3

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

$- \frac{1}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{4}$

Decimal Form:

$- 0.25$