Calculus Examples

$lim_{x \to 0} \frac{x + \cos (- 7 x) - 1}{- 2 x}$

Step 1

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

$\frac{1}{- 2} lim_{x \to 0} \frac{x + \cos (- 7 x) - 1}{x}$

Step 2

Apply L'Hospital's rule.

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Step 2.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 2.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{1}{- 2} \cdot \frac{lim_{x \to 0} x + \cos (- 7 x) - 1}{lim_{x \to 0} x}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

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

$\frac{1}{- 2} \cdot \frac{lim_{x \to 0} x + lim_{x \to 0} \cos (- 7 x) - lim_{x \to 0} 1}{lim_{x \to 0} x}$

Step 2.1.2.2

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

$\frac{1}{- 2} \cdot \frac{lim_{x \to 0} x + \cos (lim_{x \to 0} - 7 x) - lim_{x \to 0} 1}{lim_{x \to 0} x}$

Step 2.1.2.3

Move the term $- 7$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 2} \cdot \frac{lim_{x \to 0} x + \cos (- 7 lim_{x \to 0} x) - lim_{x \to 0} 1}{lim_{x \to 0} x}$

Step 2.1.2.4

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

$\frac{1}{- 2} \cdot \frac{lim_{x \to 0} x + \cos (- 7 lim_{x \to 0} x) - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.5

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

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Step 2.1.2.5.1

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

$\frac{1}{- 2} \cdot \frac{0 + \cos (- 7 lim_{x \to 0} x) - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.5.2

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

$\frac{1}{- 2} \cdot \frac{0 + \cos (- 7 \cdot 0) - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.6

Simplify the answer.

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Step 2.1.2.6.1

Simplify each term.

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Step 2.1.2.6.1.1

Multiply $- 7$ by $0$ .

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

Step 2.1.2.6.1.2

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

$\frac{1}{- 2} \cdot \frac{0 + 1 - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.6.1.3

Multiply $- 1$ by $1$ .

$\frac{1}{- 2} \cdot \frac{0 + 1 - 1}{lim_{x \to 0} x}$

Step 2.1.2.6.2

Add $0$ and $1$ .

$\frac{1}{- 2} \cdot \frac{1 - 1}{lim_{x \to 0} x}$

Step 2.1.2.6.3

Subtract $1$ from $1$ .

$\frac{1}{- 2} \cdot \frac{0}{lim_{x \to 0} x}$

Step 2.1.3

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

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

Step 2.1.4

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

Undefined

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

Step 2.2

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.

$lim_{x \to 0} \frac{x + \cos (- 7 x) - 1}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [x + \cos (- 7 x) - 1]}{\frac{d}{d x} [x]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$\frac{1}{- 2} lim_{x \to 0} \frac{\frac{d}{d x} [x + \cos (- 7 x) - 1]}{\frac{d}{d x} [x]}$

Step 2.3.2

By the Sum Rule, the derivative of $x + \cos (- 7 x) - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\cos (- 7 x)] + \frac{d}{d x} [- 1]$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{\frac{d}{d x} [x] + \frac{d}{d x} [\cos (- 7 x)] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3

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

$\frac{1}{- 2} lim_{x \to 0} \frac{1 + \frac{d}{d x} [\cos (- 7 x)] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.4

Evaluate $\frac{d}{d x} [\cos (- 7 x)]$ .

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Step 2.3.4.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = - 7 x$ .

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Step 2.3.4.1.1

To apply the Chain Rule, set $u$ as $- 7 x$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{1 + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [- 7 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.4.1.2

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

$\frac{1}{- 2} lim_{x \to 0} \frac{1 - \sin (u) \frac{d}{d x} [- 7 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.4.1.3

Replace all occurrences of $u$ with $- 7 x$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{1 - \sin (- 7 x) \frac{d}{d x} [- 7 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.4.2

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7 x$ with respect to $x$ is $- 7 \frac{d}{d x} [x]$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{1 - \sin (- 7 x) \cdot - 7 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.4.3

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

$\frac{1}{- 2} lim_{x \to 0} \frac{1 - \sin (- 7 x) \cdot - 7 \cdot 1 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.4.4

Multiply $- 7$ by $1$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{1 - \sin (- 7 x) \cdot - 7 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.4.5

Multiply $- 7$ by $- 1$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{1 + 7 \sin (- 7 x) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.5

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

$\frac{1}{- 2} lim_{x \to 0} \frac{1 + 7 \sin (- 7 x) + 0}{\frac{d}{d x} [x]}$

Step 2.3.6

Add $1 + 7 \sin (- 7 x)$ and $0$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{1 + 7 \sin (- 7 x)}{\frac{d}{d x} [x]}$

Step 2.3.7

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

$\frac{1}{- 2} lim_{x \to 0} \frac{1 + 7 \sin (- 7 x)}{1}$

Step 2.4

Divide $1 + 7 \sin (- 7 x)$ by $1$ .

$\frac{1}{- 2} lim_{x \to 0} 1 + 7 \sin (- 7 x)$

Step 3

Evaluate the limit.

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Step 3.1

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

$\frac{1}{- 2} (lim_{x \to 0} 1 + lim_{x \to 0} 7 \sin (- 7 x))$

Step 3.2

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

$\frac{1}{- 2} (1 + lim_{x \to 0} 7 \sin (- 7 x))$

Step 3.3

Move the term $7$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 2} (1 + 7 lim_{x \to 0} \sin (- 7 x))$

Step 3.4

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

$\frac{1}{- 2} (1 + 7 \sin (lim_{x \to 0} - 7 x))$

Step 3.5

Move the term $- 7$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 2} (1 + 7 \sin (- 7 lim_{x \to 0} x))$

Step 4

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

$\frac{1}{- 2} (1 + 7 \sin (- 7 \cdot 0))$

Step 5

Simplify the answer.

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Step 5.1

Move the negative in front of the fraction.

$- \frac{1}{2} (1 + 7 \sin (- 7 \cdot 0))$

Step 5.2

Simplify each term.

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Step 5.2.1

Multiply $- 7$ by $0$ .

$- \frac{1}{2} (1 + 7 \sin (0))$

Step 5.2.2

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

$- \frac{1}{2} (1 + 7 \cdot 0)$

Step 5.2.3

Multiply $7$ by $0$ .

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

Step 5.3

Add $1$ and $0$ .

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

Step 5.4

Multiply $- 1$ by $1$ .

$- \frac{1}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{2}$

Decimal Form:

$- 0.5$