Calculus Examples

$lim_{x \to 1} \frac{1 - x + \ln (x)}{1 + \cos (5 π x)}$

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.

$\frac{lim_{x \to 1} 1 - x + \ln (x)}{lim_{x \to 1} 1 + \cos (5 π x)}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{lim_{x \to 1} 1 - lim_{x \to 1} x + lim_{x \to 1} \ln (x)}{lim_{x \to 1} 1 + \cos (5 π x)}$

Step 1.1.2.2

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

$\frac{1 - lim_{x \to 1} x + lim_{x \to 1} \ln (x)}{lim_{x \to 1} 1 + \cos (5 π x)}$

Step 1.1.2.3

Move the limit inside the logarithm.

$\frac{1 - lim_{x \to 1} x + \ln (lim_{x \to 1} x)}{lim_{x \to 1} 1 + \cos (5 π x)}$

Step 1.1.2.4

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 1.1.2.4.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1 - 1 + \ln (lim_{x \to 1} x)}{lim_{x \to 1} 1 + \cos (5 π x)}$

Step 1.1.2.4.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1 - 1 + \ln (1)}{lim_{x \to 1} 1 + \cos (5 π x)}$

Step 1.1.2.5

Simplify the answer.

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Step 1.1.2.5.1

The natural logarithm of $1$ is $0$ .

$\frac{1 - 1 + 0}{lim_{x \to 1} 1 + \cos (5 π x)}$

Step 1.1.2.5.2

Subtract $1$ from $1$ .

$\frac{0 + 0}{lim_{x \to 1} 1 + \cos (5 π x)}$

Step 1.1.2.5.3

Add $0$ and $0$ .

$\frac{0}{lim_{x \to 1} 1 + \cos (5 π x)}$

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

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{0}{lim_{x \to 1} 1 + lim_{x \to 1} \cos (5 π x)}$

Step 1.1.3.1.2

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

$\frac{0}{1 + lim_{x \to 1} \cos (5 π x)}$

Step 1.1.3.1.3

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

$\frac{0}{1 + \cos (lim_{x \to 1} 5 π x)}$

Step 1.1.3.1.4

Move the term $5 π$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{1 + \cos (5 π lim_{x \to 1} x)}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{0}{1 + \cos (5 π \cdot 1)}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Multiply $5$ by $1$ .

$\frac{0}{1 + \cos (5 π)}$

Step 1.1.3.3.1.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\frac{0}{1 + \cos (π)}$

Step 1.1.3.3.1.3

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.

$\frac{0}{1 - \cos (0)}$

Step 1.1.3.3.1.4

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

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

Step 1.1.3.3.1.5

Multiply $- 1$ by $1$ .

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

Step 1.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

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.

$lim_{x \to 1} \frac{1 - x + \ln (x)}{1 + \cos (5 π x)} = lim_{x \to 1} \frac{\frac{d}{d x} [1 - x + \ln (x)]}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 1} \frac{\frac{d}{d x} [1 - x + \ln (x)]}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.2

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

$lim_{x \to 1} \frac{\frac{d}{d x} [1] + \frac{d}{d x} [- x] + \frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.3

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

$lim_{x \to 1} \frac{0 + \frac{d}{d x} [- x] + \frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- x]$ .

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Step 1.3.4.1

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

$lim_{x \to 1} \frac{0 - \frac{d}{d x} [x] + \frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.4.2

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

$lim_{x \to 1} \frac{0 - 1 \cdot 1 + \frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.4.3

Multiply $- 1$ by $1$ .

$lim_{x \to 1} \frac{0 - 1 + \frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.5

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$lim_{x \to 1} \frac{0 - 1 + \frac{1}{x}}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.6

Simplify.

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Step 1.3.6.1

Subtract $1$ from $0$ .

$lim_{x \to 1} \frac{- 1 + \frac{1}{x}}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.6.2

Reorder terms.

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{\frac{d}{d x} [1 + \cos (5 π x)]}$

Step 1.3.7

By the Sum Rule, the derivative of $1 + \cos (5 π x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\cos (5 π x)]$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{\frac{d}{d x} [1] + \frac{d}{d x} [\cos (5 π x)]}$

Step 1.3.8

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

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{0 + \frac{d}{d x} [\cos (5 π x)]}$

Step 1.3.9

Evaluate $\frac{d}{d x} [\cos (5 π x)]$ .

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Step 1.3.9.1

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 $g (x) = 5 π x$ .

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Step 1.3.9.1.1

To apply the Chain Rule, set $u$ as $5 π x$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{0 + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [5 π x]}$

Step 1.3.9.1.2

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

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{0 - \sin (u) \frac{d}{d x} [5 π x]}$

Step 1.3.9.1.3

Replace all occurrences of $u$ with $5 π x$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{0 - \sin (5 π x) \frac{d}{d x} [5 π x]}$

Step 1.3.9.2

Since $5 π$ is constant with respect to $x$ , the derivative of $5 π x$ with respect to $x$ is $5 π \frac{d}{d x} [x]$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{0 - \sin (5 π x) (5 π \frac{d}{d x} [x])}$

Step 1.3.9.3

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

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{0 - \sin (5 π x) (5 π \cdot 1)}$

Step 1.3.9.4

Multiply $5$ by $1$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{0 - \sin (5 π x) (5 π)}$

Step 1.3.9.5

Multiply $5$ by $- 1$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{0 - 5 \sin (5 π x) π}$

Step 1.3.10

Simplify.

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Step 1.3.10.1

Subtract $5 \sin (5 π x) π$ from $0$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{- 5 \sin (5 π x) π}$

Step 1.3.10.2

Reorder the factors of $- 5 \sin (5 π x) π$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1}{- 5 π \sin (5 π x)}$

Step 1.4

Combine terms.

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Step 1.4.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$lim_{x \to 1} \frac{\frac{1}{x} - 1 \cdot \frac{x}{x}}{- 5 π \sin (5 π x)}$

Step 1.4.2

Combine $- 1$ and $\frac{x}{x}$ .

$lim_{x \to 1} \frac{\frac{1}{x} + \frac{- x}{x}}{- 5 π \sin (5 π x)}$

Step 1.4.3

Combine the numerators over the common denominator.

$lim_{x \to 1} \frac{\frac{1 - x}{x}}{- 5 π \sin (5 π x)}$

Step 2

Evaluate the limit.

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Step 2.1

Move the term $\frac{1}{- 5 π}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{\frac{1 - x}{x}}{\sin (5 π x)}$

Step 2.2

Simplify the limit argument.

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Step 2.2.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{- 5 π} lim_{x \to 1} \frac{1 - x}{x} \cdot \frac{1}{\sin (5 π x)}$

Step 2.2.2

Multiply $\frac{1 - x}{x}$ by $\frac{1}{\sin (5 π x)}$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{1 - x}{x \sin (5 π x)}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

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

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Step 3.1.1

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

$\frac{1}{- 5 π} \cdot \frac{lim_{x \to 1} 1 - x}{lim_{x \to 1} x \sin (5 π x)}$

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{1}{- 5 π} \cdot \frac{lim_{x \to 1} 1 - lim_{x \to 1} x}{lim_{x \to 1} x \sin (5 π x)}$

Step 3.1.2.2

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

$\frac{1}{- 5 π} \cdot \frac{1 - lim_{x \to 1} x}{lim_{x \to 1} x \sin (5 π x)}$

Step 3.1.2.3

Simplify the expression.

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Step 3.1.2.3.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1}{- 5 π} \cdot \frac{1 - 1}{lim_{x \to 1} x \sin (5 π x)}$

Step 3.1.2.3.2

Subtract $1$ from $1$ .

$\frac{1}{- 5 π} \cdot \frac{0}{lim_{x \to 1} x \sin (5 π x)}$

Step 3.1.3

Evaluate the limit of the denominator.

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Step 3.1.3.1

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{1}{- 5 π} \cdot \frac{0}{lim_{x \to 1} x \cdot lim_{x \to 1} \sin (5 π x)}$

Step 3.1.3.2

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

$\frac{1}{- 5 π} \cdot \frac{0}{lim_{x \to 1} x \cdot \sin (lim_{x \to 1} 5 π x)}$

Step 3.1.3.3

Move the term $5 π$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 5 π} \cdot \frac{0}{lim_{x \to 1} x \cdot \sin (5 π lim_{x \to 1} x)}$

Step 3.1.3.4

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 3.1.3.4.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1}{- 5 π} \cdot \frac{0}{1 \cdot \sin (5 π lim_{x \to 1} x)}$

Step 3.1.3.4.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1}{- 5 π} \cdot \frac{0}{1 \cdot \sin (5 π \cdot 1)}$

Step 3.1.3.5

Simplify the answer.

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Step 3.1.3.5.1

Multiply $\sin (5 π \cdot 1)$ by $1$ .

$\frac{1}{- 5 π} \cdot \frac{0}{\sin (5 π \cdot 1)}$

Step 3.1.3.5.2

Multiply $5$ by $1$ .

$\frac{1}{- 5 π} \cdot \frac{0}{\sin (5 π)}$

Step 3.1.3.5.3

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\frac{1}{- 5 π} \cdot \frac{0}{\sin (π)}$

Step 3.1.3.5.4

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

$\frac{1}{- 5 π} \cdot \frac{0}{\sin (0)}$

Step 3.1.3.5.5

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

$\frac{1}{- 5 π} \cdot \frac{0}{0}$

Step 3.1.3.5.6

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

Undefined

$\frac{1}{- 5 π} \cdot \frac{0}{0}$

Step 3.1.3.6

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

Undefined

$\frac{1}{- 5 π} \cdot \frac{0}{0}$

Step 3.1.4

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

Undefined

$\frac{1}{- 5 π} \cdot \frac{0}{0}$

Step 3.2

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.

$lim_{x \to 1} \frac{1 - x}{x \sin (5 π x)} = lim_{x \to 1} \frac{\frac{d}{d x} [1 - x]}{\frac{d}{d x} [x \sin (5 π x)]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$\frac{1}{- 5 π} lim_{x \to 1} \frac{\frac{d}{d x} [1 - x]}{\frac{d}{d x} [x \sin (5 π x)]}$

Step 3.3.2

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

$\frac{1}{- 5 π} lim_{x \to 1} \frac{\frac{d}{d x} [1] + \frac{d}{d x} [- x]}{\frac{d}{d x} [x \sin (5 π x)]}$

Step 3.3.3

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

$\frac{1}{- 5 π} lim_{x \to 1} \frac{0 + \frac{d}{d x} [- x]}{\frac{d}{d x} [x \sin (5 π x)]}$

Step 3.3.4

Evaluate $\frac{d}{d x} [- x]$ .

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Step 3.3.4.1

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

$\frac{1}{- 5 π} lim_{x \to 1} \frac{0 - \frac{d}{d x} [x]}{\frac{d}{d x} [x \sin (5 π x)]}$

Step 3.3.4.2

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

$\frac{1}{- 5 π} lim_{x \to 1} \frac{0 - 1 \cdot 1}{\frac{d}{d x} [x \sin (5 π x)]}$

Step 3.3.4.3

Multiply $- 1$ by $1$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{0 - 1}{\frac{d}{d x} [x \sin (5 π x)]}$

Step 3.3.5

Subtract $1$ from $0$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{\frac{d}{d x} [x \sin (5 π x)]}$

Step 3.3.6

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \sin (5 π x)$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \frac{d}{d x} [\sin (5 π x)] + \sin (5 π x) \frac{d}{d x} [x]}$

Step 3.3.7

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 5 π x$ .

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Step 3.3.7.1

To apply the Chain Rule, set $u$ as $5 π x$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \frac{d}{d u} [\sin (u)] \frac{d}{d x} [5 π x] + \sin (5 π x) \frac{d}{d x} [x]}$

Step 3.3.7.2

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

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \cos (u) \frac{d}{d x} [5 π x] + \sin (5 π x) \frac{d}{d x} [x]}$

Step 3.3.7.3

Replace all occurrences of $u$ with $5 π x$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \cos (5 π x) \frac{d}{d x} [5 π x] + \sin (5 π x) \frac{d}{d x} [x]}$

Step 3.3.8

Since $5 π$ is constant with respect to $x$ , the derivative of $5 π x$ with respect to $x$ is $5 π \frac{d}{d x} [x]$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \cos (5 π x) \cdot 5 π \frac{d}{d x} [x] + \sin (5 π x) \frac{d}{d x} [x]}$

Step 3.3.9

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

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \cos (5 π x) \cdot 5 π \cdot 1 + \sin (5 π x) \frac{d}{d x} [x]}$

Step 3.3.10

Multiply $5$ by $1$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \cos (5 π x) \cdot 5 π + \sin (5 π x) \frac{d}{d x} [x]}$

Step 3.3.11

Move $5$ to the left of $\cos (5 π x)$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \cdot 5 \cdot \cos (5 π x) π + \sin (5 π x) \frac{d}{d x} [x]}$

Step 3.3.12

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

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \cdot 5 \cos (5 π x) π + \sin (5 π x) \cdot 1}$

Step 3.3.13

Multiply $\sin (5 π x)$ by $1$ .

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{x \cdot 5 \cos (5 π x) π + \sin (5 π x)}$

Step 3.3.14

Reorder terms.

$\frac{1}{- 5 π} lim_{x \to 1} \frac{- 1}{5 π x \cos (5 π x) + \sin (5 π x)}$

Step 4

Evaluate the limit.

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Step 4.1

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $1$ .

$\frac{1}{- 5 π} \cdot \frac{lim_{x \to 1} - 1}{lim_{x \to 1} 5 π x \cos (5 π x) + \sin (5 π x)}$

Step 4.2

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

$\frac{1}{- 5 π} \cdot \frac{- 1}{lim_{x \to 1} 5 π x \cos (5 π x) + \sin (5 π x)}$

Step 4.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{1}{- 5 π} \cdot \frac{- 1}{lim_{x \to 1} 5 π x \cos (5 π x) + lim_{x \to 1} \sin (5 π x)}$

Step 4.4

Move the term $5 π$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π lim_{x \to 1} x \cos (5 π x) + lim_{x \to 1} \sin (5 π x)}$

Step 4.5

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π lim_{x \to 1} x \cdot lim_{x \to 1} \cos (5 π x) + lim_{x \to 1} \sin (5 π x)}$

Step 4.6

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

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π lim_{x \to 1} x \cdot \cos (lim_{x \to 1} 5 π x) + lim_{x \to 1} \sin (5 π x)}$

Step 4.7

Move the term $5 π$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π lim_{x \to 1} x \cdot \cos (5 π lim_{x \to 1} x) + lim_{x \to 1} \sin (5 π x)}$

Step 4.8

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

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π lim_{x \to 1} x \cdot \cos (5 π lim_{x \to 1} x) + \sin (lim_{x \to 1} 5 π x)}$

Step 4.9

Move the term $5 π$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π lim_{x \to 1} x \cdot \cos (5 π lim_{x \to 1} x) + \sin (5 π lim_{x \to 1} x)}$

Step 5

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 5.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π \cdot 1 \cdot \cos (5 π lim_{x \to 1} x) + \sin (5 π lim_{x \to 1} x)}$

Step 5.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π \cdot 1 \cdot \cos (5 π \cdot 1) + \sin (5 π lim_{x \to 1} x)}$

Step 5.3

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1}{- 5 π} \cdot \frac{- 1}{5 π \cdot 1 \cdot \cos (5 π \cdot 1) + \sin (5 π \cdot 1)}$

Step 6

Simplify the answer.

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Step 6.1

Move the negative in front of the fraction.

$- \frac{1}{5 π} \cdot \frac{- 1}{5 π \cdot 1 \cdot \cos (5 π \cdot 1) + \sin (5 π \cdot 1)}$

Step 6.2

Simplify the denominator.

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Step 6.2.1

Multiply $5$ by $1$ .

$- \frac{1}{5 π} \cdot \frac{- 1}{5 π \cdot \cos (5 π \cdot 1) + \sin (5 π \cdot 1)}$

Step 6.2.2

Multiply $5$ by $1$ .

$- \frac{1}{5 π} \cdot \frac{- 1}{5 π \cdot \cos (5 π) + \sin (5 π \cdot 1)}$

Step 6.2.3

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$- \frac{1}{5 π} \cdot \frac{- 1}{5 π \cdot \cos (π) + \sin (5 π \cdot 1)}$

Step 6.2.4

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.

$- \frac{1}{5 π} \cdot \frac{- 1}{5 π \cdot (- \cos (0)) + \sin (5 π \cdot 1)}$

Step 6.2.5

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

$- \frac{1}{5 π} \cdot \frac{- 1}{5 π \cdot (- 1 \cdot 1) + \sin (5 π \cdot 1)}$

Step 6.2.6

Multiply $- 1$ by $1$ .

$- \frac{1}{5 π} \cdot \frac{- 1}{5 π \cdot - 1 + \sin (5 π \cdot 1)}$

Step 6.2.7

Multiply $- 1$ by $5$ .

$- \frac{1}{5 π} \cdot \frac{- 1}{- 5 π + \sin (5 π \cdot 1)}$

Step 6.2.8

Multiply $5$ by $1$ .

$- \frac{1}{5 π} \cdot \frac{- 1}{- 5 π + \sin (5 π)}$

Step 6.2.9

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$- \frac{1}{5 π} \cdot \frac{- 1}{- 5 π + \sin (π)}$

Step 6.2.10

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

$- \frac{1}{5 π} \cdot \frac{- 1}{- 5 π + \sin (0)}$

Step 6.2.11

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

$- \frac{1}{5 π} \cdot \frac{- 1}{- 5 π + 0}$

Step 6.2.12

Add $- 5 π$ and $0$ .

$- \frac{1}{5 π} \cdot \frac{- 1}{- 5 π}$

Step 6.3

Dividing two negative values results in a positive value.

$- \frac{1}{5 π} \cdot \frac{1}{5 π}$

Step 6.4

Multiply $- \frac{1}{5 π} \cdot \frac{1}{5 π}$ .

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Step 6.4.1

Multiply $\frac{1}{5 π}$ by $\frac{1}{5 π}$ .

$- \frac{1}{5 π (5 π)}$

Step 6.4.2

Multiply $5$ by $5$ .

$- \frac{1}{25 π π}$

Step 6.4.3

Raise $π$ to the power of $1$ .

$- \frac{1}{25 (π^{1} π)}$

Step 6.4.4

Raise $π$ to the power of $1$ .

$- \frac{1}{25 (π^{1} π^{1})}$

Step 6.4.5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{1}{25 π^{1 + 1}}$

Step 6.4.6

Add $1$ and $1$ .

$- \frac{1}{25 π^{2}}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{25 π^{2}}$

Decimal Form:

$- 0.00405284 \dots$