Calculus Examples

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

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.

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Multiply $- 1$ by $0$ .

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

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

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

Step 1.2.6

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

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

Step 1.2.7

Simplify terms.

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Step 1.2.7.1

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

$\frac{\cos (\frac{π}{2} + 0) + 0}{lim_{h \to 0} h}$

Step 1.2.7.2

Simplify the answer.

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Step 1.2.7.2.1

Simplify each term.

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Step 1.2.7.2.1.1

Add $\frac{π}{2}$ and $0$ .

$\frac{\cos (\frac{π}{2}) + 0}{lim_{h \to 0} h}$

Step 1.2.7.2.1.2

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

$\frac{0 + 0}{lim_{h \to 0} h}$

Step 1.2.7.2.2

Add $0$ and $0$ .

$\frac{0}{lim_{h \to 0} h}$

Step 1.3

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

$\frac{0}{0}$

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.

$lim_{h \to 0} \frac{\cos (\frac{π}{2} - h) - 0}{h} = lim_{h \to 0} \frac{\frac{d}{d h} [\cos (\frac{π}{2} - h) - 0]}{\frac{d}{d h} [h]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

By the Sum Rule, the derivative of $\cos (\frac{π}{2} - h) - 0$ with respect to $h$ is $\frac{d}{d h} [\cos (\frac{π}{2} - h)] + \frac{d}{d h} [- 0]$ .

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

Step 3.3

Evaluate $\frac{d}{d h} [\cos (\frac{π}{2} - h)]$ .

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Step 3.3.1

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

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Step 3.3.1.1

To apply the Chain Rule, set $u$ as $\frac{π}{2} - h$ .

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

Step 3.3.1.2

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

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

Step 3.3.1.3

Replace all occurrences of $u$ with $\frac{π}{2} - h$ .

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

Step 3.3.2

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

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

Step 3.3.3

Since $\frac{π}{2}$ is constant with respect to $h$ , the derivative of $\frac{π}{2}$ with respect to $h$ is $0$ .

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

Step 3.3.4

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

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

Step 3.3.5

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

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

Step 3.3.6

Multiply $- 1$ by $1$ .

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

Step 3.3.7

Subtract $1$ from $0$ .

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

Step 3.3.8

Multiply $- 1$ by $- 1$ .

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

Step 3.3.9

Multiply $\sin (\frac{π}{2} - h)$ by $1$ .

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

Step 3.4

Evaluate $\frac{d}{d h} [- 0]$ .

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Step 3.4.1

Multiply $- 1$ by $0$ .

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

Step 3.4.2

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

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

Step 3.5

Add $\sin (\frac{π}{2} - h)$ and $0$ .

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

Step 3.6

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

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

Step 4

Combine terms.

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Step 4.1

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

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

Step 4.2

Combine $- h$ and $\frac{2}{2}$ .

$lim_{h \to 0} \frac{\sin (\frac{π}{2} + \frac{- h \cdot 2}{2})}{1}$

Step 4.3

Combine the numerators over the common denominator.

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

Step 5

Evaluate the limit.

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Step 5.1

Divide $\sin (\frac{π - h \cdot 2}{2})$ by $1$ .

$lim_{h \to 0} \sin (\frac{π - h \cdot 2}{2})$

Step 5.2

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

$\sin (lim_{h \to 0} \frac{π - h \cdot 2}{2})$

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Evaluate the limit of $π$ which is constant as $h$ approaches $0$ .

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

Step 5.6

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

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

Step 6

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

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

Step 7

Simplify the answer.

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Step 7.1

Multiply $- (2 \cdot 0)$ .

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Step 7.1.1

Multiply $2$ by $0$ .

$\sin (\frac{1}{2} (π - 0))$

Step 7.1.2

Multiply $- 1$ by $0$ .

$\sin (\frac{1}{2} (π + 0))$

Step 7.2

Add $π$ and $0$ .

$\sin (\frac{1}{2} π)$

Step 7.3

Combine $\frac{1}{2}$ and $π$ .

$\sin (\frac{π}{2})$

Step 7.4

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

$1$