Calculus Examples

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

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.

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Move the limit inside the logarithm.

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

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

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

Step 1.3.1.2

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

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

Step 1.3.2

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

$\frac{0}{\sin (5 π \cdot 1)}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Multiply $5$ by $1$ .

$\frac{0}{\sin (5 π)}$

Step 1.3.3.2

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

$\frac{0}{\sin (π)}$

Step 1.3.3.3

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

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

Step 1.3.3.4

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

$\frac{0}{0}$

Step 1.3.3.5

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

Undefined

$\frac{0}{0}$

Step 1.3.4

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

Step 3.3

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 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

Multiply $5$ by $1$ .

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

Step 3.7

Remove parentheses.

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

Step 3.8

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

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

Step 3.9

Multiply $5$ by $\cos (5 π x)$ .

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

Step 3.10

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

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

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

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

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

Step 6

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

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

Step 7

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

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

Step 8

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

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

Step 9

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

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

Step 10

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

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

Step 11

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

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

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 $x$ .

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

Step 12.2

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

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

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.

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

Step 13.1.2

Rewrite the expression.

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

Step 13.2

Convert from $\frac{1}{\cos (5 π \cdot 1)}$ to $\sec (5 π \cdot 1)$ .

$\frac{1}{5 π} \sec (5 π \cdot 1)$

Step 13.3

Multiply $5 π$ by $1$ .

$\frac{1}{5 π} \sec (5 π)$

Step 13.4

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

$\frac{1}{5 π} \sec (π)$

Step 13.5

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

$\frac{1}{5 π} (- \sec (0))$

Step 13.6

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

$\frac{1}{5 π} (- 1 \cdot 1)$

Step 13.7

Multiply $- 1$ by $1$ .

$\frac{1}{5 π} \cdot - 1$

Step 13.8

Combine $\frac{1}{5 π}$ and $- 1$ .

$\frac{- 1}{5 π}$

Step 13.9

Move the negative in front of the fraction.

$- \frac{1}{5 π}$