Calculus Examples

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

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.

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Evaluate the limit.

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Step 1.2.1.1

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

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

Step 1.2.1.2

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

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

Step 1.2.2

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

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

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Multiply $- 1$ by $1$ .

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

Step 1.2.3.2

Subtract $1$ from $1$ .

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

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

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

Step 1.3.2

Move the limit inside the logarithm.

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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Step 1.3.5.1

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

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

Step 1.3.5.2

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

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

Step 1.3.6

Simplify the answer.

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Step 1.3.6.1

Simplify each term.

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Step 1.3.6.1.1

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

$\frac{0}{0 - \sin (π \cdot 1)}$

Step 1.3.6.1.2

Multiply $π$ by $1$ .

$\frac{0}{0 - \sin (π)}$

Step 1.3.6.1.3

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

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

Step 1.3.6.1.4

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

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

Step 1.3.6.1.5

Multiply $- 1$ by $0$ .

$\frac{0}{0 + 0}$

Step 1.3.6.2

Add $0$ and $0$ .

$\frac{0}{0}$

Step 1.3.6.3

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

Undefined

$\frac{0}{0}$

Step 1.3.7

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

Undefined

$\frac{0}{0}$

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Add $1$ and $0$ .

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

Step 3.6

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

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

Step 3.7

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

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

Step 3.8

Evaluate $\frac{d}{d x} [- \sin (π x)]$ .

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Step 3.8.1

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

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

Step 3.8.2

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 $g (x) = π x$ .

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Step 3.8.2.1

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

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

Step 3.8.2.2

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

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

Step 3.8.2.3

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

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

Step 3.8.3

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

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

Step 3.8.4

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

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

Step 3.8.5

Multiply $π$ by $1$ .

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

Step 3.8.6

Remove parentheses.

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

Step 3.9

Reorder terms.

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

Step 4

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

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

Step 5

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

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

Step 6

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

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

Step 7

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

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

Step 8

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

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

Step 9

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

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

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

Step 12.2

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

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

Step 13

Simplify the answer.

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Step 13.1

Cancel the common factor of $1$ .

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Step 13.1.1

Cancel the common factor.

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

Step 13.1.2

Rewrite the expression.

$\frac{1}{- π \cos (π \cdot 1) + 1}$

Step 13.2

Simplify the denominator.

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Step 13.2.1

Multiply $π$ by $1$ .

$\frac{1}{- π \cos (π) + 1}$

Step 13.2.2

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.

$\frac{1}{- π (- \cos (0)) + 1}$

Step 13.2.3

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

$\frac{1}{- π (- 1 \cdot 1) + 1}$

Step 13.2.4

Multiply $- 1$ by $1$ .

$\frac{1}{- π \cdot - 1 + 1}$

Step 13.2.5

Multiply $- π \cdot - 1$ .

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Step 13.2.5.1

Multiply $- 1$ by $- 1$ .

$\frac{1}{1 π + 1}$

Step 13.2.5.2

Multiply $π$ by $1$ .

$\frac{1}{π + 1}$