Calculus Examples

$lim_{h \to 0} \frac{\cos (π + h) + 1}{h}$

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.

$\frac{lim_{h \to 0} \cos (π + h) + 1}{lim_{h \to 0} h}$

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 $h$ approaches $0$ .

$\frac{lim_{h \to 0} \cos (π + h) + lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 1.1.2.1.2

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

$\frac{\cos (lim_{h \to 0} π + h) + lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 1.1.2.1.3

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

$\frac{\cos (lim_{h \to 0} π + lim_{h \to 0} h) + lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 1.1.2.1.4

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

$\frac{\cos (π + lim_{h \to 0} h) + lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 1.1.2.1.5

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

$\frac{\cos (π + lim_{h \to 0} h) + 1}{lim_{h \to 0} h}$

Step 1.1.2.2

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

$\frac{\cos (π + 0) + 1}{lim_{h \to 0} h}$

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

Add $π$ and $0$ .

$\frac{\cos (π) + 1}{lim_{h \to 0} h}$

Step 1.1.2.3.1.2

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

$\frac{- \cos (0) + 1}{lim_{h \to 0} h}$

Step 1.1.2.3.1.3

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

$\frac{- 1 \cdot 1 + 1}{lim_{h \to 0} h}$

Step 1.1.2.3.1.4

Multiply $- 1$ by $1$ .

$\frac{- 1 + 1}{lim_{h \to 0} h}$

Step 1.1.2.3.2

Add $- 1$ and $1$ .

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

Step 1.1.3

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

$\frac{0}{0}$

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.

$lim_{h \to 0} \frac{\cos (π + h) + 1}{h} = lim_{h \to 0} \frac{\frac{d}{d h} [\cos (π + h) + 1]}{\frac{d}{d h} [h]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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Step 1.3.3.1

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 $g (h) = π + h$ .

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Step 1.3.3.1.1

To apply the Chain Rule, set $u$ as $π + h$ .

$lim_{h \to 0} \frac{\frac{d}{d u} [\cos (u)] \frac{d}{d h} [π + h] + \frac{d}{d h} [1]}{\frac{d}{d h} [h]}$

Step 1.3.3.1.2

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

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

Step 1.3.3.1.3

Replace all occurrences of $u$ with $π + h$ .

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

Step 1.3.3.2

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

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

Step 1.3.3.3

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

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

Step 1.3.3.4

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

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

Step 1.3.3.5

Add $0$ and $1$ .

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

Step 1.3.3.6

Multiply $- 1$ by $1$ .

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

Step 1.3.4

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

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

Step 1.3.5

Add $- \sin (π + h)$ and $0$ .

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

Step 1.3.6

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

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

Step 1.4

Divide $- \sin (π + h)$ by $1$ .

$lim_{h \to 0} - \sin (π + h)$

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 $h$ .

$- lim_{h \to 0} \sin (π + h)$

Step 2.2

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

$- \sin (lim_{h \to 0} π + h)$

Step 2.3

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

$- \sin (lim_{h \to 0} π + lim_{h \to 0} h)$

Step 2.4

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

$- \sin (π + lim_{h \to 0} h)$

Step 3

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

$- \sin (π + 0)$

Step 4

Simplify the answer.

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Step 4.1

Add $π$ and $0$ .

$- \sin (π)$

Step 4.2

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

$- \sin (0)$

Step 4.3

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

$- 0$

Step 4.4

Multiply $- 1$ by $0$ .

$0$