Calculus Examples

$lim_{x \to 0} \frac{\ln (1 - x) - \sin (x)}{1 - \cos^{2} (x)}$

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.

$\frac{lim_{x \to 0} \ln (1 - x) - \sin (x)}{lim_{x \to 0} 1 - \cos^{2} (x)}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

$\frac{lim_{x \to 0} \ln (1 - x) - lim_{x \to 0} \sin (x)}{lim_{x \to 0} 1 - \cos^{2} (x)}$

Step 1.2.2

Move the limit inside the logarithm.

$\frac{\ln (lim_{x \to 0} 1 - x) - lim_{x \to 0} \sin (x)}{lim_{x \to 0} 1 - \cos^{2} (x)}$

Step 1.2.3

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

$\frac{\ln (lim_{x \to 0} 1 - lim_{x \to 0} x) - lim_{x \to 0} \sin (x)}{lim_{x \to 0} 1 - \cos^{2} (x)}$

Step 1.2.4

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

$\frac{\ln (1 - lim_{x \to 0} x) - lim_{x \to 0} \sin (x)}{lim_{x \to 0} 1 - \cos^{2} (x)}$

Step 1.2.5

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

$\frac{\ln (1 - lim_{x \to 0} x) - \sin (lim_{x \to 0} x)}{lim_{x \to 0} 1 - \cos^{2} (x)}$

Step 1.2.6

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

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Step 1.2.6.1

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

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

Step 1.2.6.2

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

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

Step 1.2.7

Simplify the answer.

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Step 1.2.7.1

Simplify each term.

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Step 1.2.7.1.1

Add $1$ and $0$ .

$\frac{\ln (1) - \sin (0)}{lim_{x \to 0} 1 - \cos^{2} (x)}$

Step 1.2.7.1.2

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

$\frac{0 - \sin (0)}{lim_{x \to 0} 1 - \cos^{2} (x)}$

Step 1.2.7.1.3

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

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

Step 1.2.7.1.4

Multiply $- 1$ by $0$ .

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

Step 1.2.7.2

Add $0$ and $0$ .

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

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

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

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

Step 1.3.1.2

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

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

Step 1.3.1.3

Move the exponent $2$ from $\cos^{2} (x)$ outside the limit using the Limits Power Rule.

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

Step 1.3.1.4

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

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

Step 1.3.2

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

$\frac{0}{1 - \cos^{2} (0)}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Apply pythagorean identity.

$\frac{0}{\sin^{2} (0)}$

Step 1.3.3.2

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

$\frac{0}{0^{2}}$

Step 1.3.3.3

Raising $0$ to any positive power yields $0$ .

$\frac{0}{0}$

Step 1.3.3.4

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.

$lim_{x \to 0} \frac{\ln (1 - x) - \sin (x)}{1 - \cos^{2} (x)} = lim_{x \to 0} \frac{\frac{d}{d x} [\ln (1 - x) - \sin (x)]}{\frac{d}{d x} [1 - \cos^{2} (x)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [\ln (1 - x) - \sin (x)]}{\frac{d}{d x} [1 - \cos^{2} (x)]}$

Step 3.2

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

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

Step 3.3

Evaluate $\frac{d}{d x} [\ln (1 - x)]$ .

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Step 3.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 1 - x$ .

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Step 3.3.1.1

To apply the Chain Rule, set $u$ as $1 - x$ .

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

Step 3.3.1.2

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

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

Step 3.3.1.3

Replace all occurrences of $u$ with $1 - x$ .

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

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

Step 3.3.5

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

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

Step 3.3.6

Multiply $- 1$ by $1$ .

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

Step 3.3.7

Subtract $1$ from $0$ .

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

Step 3.3.8

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

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

Step 3.3.9

Move the negative in front of the fraction.

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.5

By the Sum Rule, the derivative of $1 - \cos^{2} (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \cos^{2} (x)]$ .

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

Step 3.6

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

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

Step 3.7

Evaluate $\frac{d}{d x} [- \cos^{2} (x)]$ .

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Step 3.7.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos^{2} (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos^{2} (x)]$ .

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

Step 3.7.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \cos (x)$ .

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Step 3.7.2.1

To apply the Chain Rule, set $u$ as $\cos (x)$ .

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

Step 3.7.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

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

Step 3.7.2.3

Replace all occurrences of $u$ with $\cos (x)$ .

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

Step 3.7.3

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

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

Step 3.7.4

Multiply $- 1$ by $2$ .

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

Step 3.7.5

Multiply $- 2$ by $- 1$ .

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

Step 3.7.6

Remove parentheses.

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

Step 3.8

Simplify.

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Step 3.8.1

Add $0$ and $2 \cos (x) \sin (x)$ .

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

Step 3.8.2

Reorder $2 \cos (x)$ and $\sin (x)$ .

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

Step 3.8.3

Reorder $\sin (x)$ and $2$ .

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

Step 3.8.4

Apply the sine double-angle identity.

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

Step 4

Since the function approaches $\infty$ from the left and $- \infty$ from the right, the limit does not exist.

$Does not exist$