Calculus Examples

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

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${(1 - \cos (x))}^{x}$ as $e^{\ln ({(1 - \cos (x))}^{x})}$ .

$lim_{x \to 0} e^{\ln ({(1 - \cos (x))}^{x})}$

Step 1.2

Expand $\ln ({(1 - \cos (x))}^{x})$ by moving $x$ outside the logarithm.

$lim_{x \to 0} e^{x \ln (1 - \cos (x))}$

Step 2

Move the limit into the exponent.

$e^{lim_{x \to 0} x \ln (1 - \cos (x))}$

Step 3

Rewrite $x \ln (1 - \cos (x))$ as $\frac{\ln (1 - \cos (x))}{\frac{1}{x}}$ .

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

Step 4

Set up the limit as a left-sided limit.

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

Step 5

Evaluate the left-sided limit.

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Step 5.1

Apply L'Hospital's rule.

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Step 5.1.1

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

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Step 5.1.1.1

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

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

Step 5.1.1.2

As $x$ approaches $0$ from the left side, $\ln (1 - \cos (x))$ decreases without bound.

$e^{\frac{- \infty}{lim_{x \to 0^{-}} \frac{1}{x}}}$

Step 5.1.1.3

Since the numerator is a constant and the denominator approaches $0$ when $x$ approaches $0$ from the left, the fraction $\frac{1}{x}$ approaches negative infinity.

$e^{\frac{- \infty}{- \infty}}$

Step 5.1.1.4

Infinity divided by infinity is undefined.

Undefined

$e^{\frac{- \infty}{- \infty}}$

Step 5.1.2

Since $\frac{- \infty}{- \infty}$ 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.

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

Step 5.1.3

Find the derivative of the numerator and denominator.

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Step 5.1.3.1

Differentiate the numerator and denominator.

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

Step 5.1.3.2

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

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Step 5.1.3.2.1

To apply the Chain Rule, set $u$ as $1 - \cos (x)$ .

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

Step 5.1.3.2.2

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

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

Step 5.1.3.2.3

Replace all occurrences of $u$ with $1 - \cos (x)$ .

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

Step 5.1.3.3

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

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

Step 5.1.3.4

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

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

Step 5.1.3.5

Add $0$ and $\frac{d}{d x} [- \cos (x)]$ .

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

Step 5.1.3.6

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

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

Step 5.1.3.7

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

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

Step 5.1.3.8

Multiply $- 1$ by $- 1$ .

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

Step 5.1.3.9

Multiply $\sin (x)$ by $1$ .

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

Step 5.1.3.10

Combine $\frac{1}{1 - \cos (x)}$ and $\sin (x)$ .

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

Step 5.1.3.11

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 5.1.3.12

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

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

Step 5.1.3.13

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 5.1.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.1.5

Combine $\frac{\sin (x)}{1 - \cos (x)}$ and $x^{2}$ .

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

Step 5.1.6

Reorder factors in $- \frac{\sin (x) x^{2}}{1 - \cos (x)}$ .

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

Step 5.2

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

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

Step 5.3

Apply L'Hospital's rule.

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Step 5.3.1

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

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Step 5.3.1.1

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

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

Step 5.3.1.2

Evaluate the limit of the numerator.

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Step 5.3.1.2.1

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

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

Step 5.3.1.2.2

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

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

Step 5.3.1.2.3

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

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

Step 5.3.1.2.4

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

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Step 5.3.1.2.4.1

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

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

Step 5.3.1.2.4.2

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

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

Step 5.3.1.2.5

Simplify the answer.

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Step 5.3.1.2.5.1

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

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

Step 5.3.1.2.5.2

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

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

Step 5.3.1.2.5.3

Multiply $0$ by $0$ .

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

Step 5.3.1.3

Evaluate the limit of the denominator.

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Step 5.3.1.3.1

Evaluate the limit.

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Step 5.3.1.3.1.1

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

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

Step 5.3.1.3.1.2

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

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

Step 5.3.1.3.1.3

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

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

Step 5.3.1.3.2

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

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

Step 5.3.1.3.3

Simplify the answer.

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Step 5.3.1.3.3.1

Simplify each term.

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Step 5.3.1.3.3.1.1

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

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

Step 5.3.1.3.3.1.2

Multiply $- 1$ by $1$ .

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

Step 5.3.1.3.3.2

Subtract $1$ from $1$ .

$e^{- \frac{0}{0}}$

Step 5.3.1.3.3.3

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

Undefined

$e^{- \frac{0}{0}}$

Step 5.3.1.3.4

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

Undefined

$e^{- \frac{0}{0}}$

Step 5.3.1.4

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

Undefined

$e^{- \frac{0}{0}}$

Step 5.3.2

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.

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

Step 5.3.3

Find the derivative of the numerator and denominator.

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Step 5.3.3.1

Differentiate the numerator and denominator.

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

Step 5.3.3.2

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 $f (x) = x^{2}$ and $g (x) = \sin (x)$ .

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

Step 5.3.3.3

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

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

Step 5.3.3.4

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

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

Step 5.3.3.5

Reorder terms.

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

Step 5.3.3.6

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

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

Step 5.3.3.7

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

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

Step 5.3.3.8

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

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Step 5.3.3.8.1

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

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

Step 5.3.3.8.2

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

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

Step 5.3.3.8.3

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.8.4

Multiply $\sin (x)$ by $1$ .

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

Step 5.3.3.9

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

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

Step 5.4

Since $\frac{x^{2} \cdot - 1 + 2 x \sin (x)}{\sin (x)} \leq \frac{x^{2} \cos (x) + 2 x \sin (x)}{\sin (x)} \leq \frac{x^{2} \cdot 1 + 2 x \sin (x)}{\sin (x)}$ and $lim_{x \to 0^{-}} \frac{x^{2} \cdot - 1 + 2 x \sin (x)}{\sin (x)} = lim_{x \to 0^{-}} \frac{x^{2} \cdot 1 + 2 x \sin (x)}{\sin (x)} = 0$ , apply the squeeze theorem.

$e^{- 0}$

Step 5.5

Multiply $- 1$ by $0$ .

$e^{0}$

Step 6

Set up the limit as a right-sided limit.

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

Step 7

Evaluate the right-sided limit.

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Step 7.1

Apply L'Hospital's rule.

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Step 7.1.1

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

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Step 7.1.1.1

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

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

Step 7.1.1.2

As $x$ approaches $0$ from the right side, $\ln (1 - \cos (x))$ decreases without bound.

$e^{\frac{- \infty}{lim_{x \to 0^{+}} \frac{1}{x}}}$

Step 7.1.1.3

Since the numerator is a constant and the denominator approaches $0$ when $x$ approaches $0$ from the right, the fraction $\frac{1}{x}$ approaches infinity.

$e^{\frac{- \infty}{\infty}}$

Step 7.1.1.4

Infinity divided by infinity is undefined.

Undefined

$e^{\frac{- \infty}{\infty}}$

Step 7.1.2

Since $\frac{- \infty}{\infty}$ 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.

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

Step 7.1.3

Find the derivative of the numerator and denominator.

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Step 7.1.3.1

Differentiate the numerator and denominator.

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

Step 7.1.3.2

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

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Step 7.1.3.2.1

To apply the Chain Rule, set $u$ as $1 - \cos (x)$ .

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

Step 7.1.3.2.2

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

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

Step 7.1.3.2.3

Replace all occurrences of $u$ with $1 - \cos (x)$ .

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

Step 7.1.3.3

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

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

Step 7.1.3.4

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

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

Step 7.1.3.5

Add $0$ and $\frac{d}{d x} [- \cos (x)]$ .

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

Step 7.1.3.6

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

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

Step 7.1.3.7

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

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

Step 7.1.3.8

Multiply $- 1$ by $- 1$ .

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

Step 7.1.3.9

Multiply $\sin (x)$ by $1$ .

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

Step 7.1.3.10

Combine $\frac{1}{1 - \cos (x)}$ and $\sin (x)$ .

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

Step 7.1.3.11

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 7.1.3.12

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

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

Step 7.1.3.13

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 7.1.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 7.1.5

Combine $\frac{\sin (x)}{1 - \cos (x)}$ and $x^{2}$ .

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

Step 7.1.6

Reorder factors in $- \frac{\sin (x) x^{2}}{1 - \cos (x)}$ .

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

Step 7.2

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

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

Step 7.3

Apply L'Hospital's rule.

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Step 7.3.1

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

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Step 7.3.1.1

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

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

Step 7.3.1.2

Evaluate the limit of the numerator.

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Step 7.3.1.2.1

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

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

Step 7.3.1.2.2

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

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

Step 7.3.1.2.3

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

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

Step 7.3.1.2.4

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

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Step 7.3.1.2.4.1

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

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

Step 7.3.1.2.4.2

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

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

Step 7.3.1.2.5

Simplify the answer.

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Step 7.3.1.2.5.1

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

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

Step 7.3.1.2.5.2

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

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

Step 7.3.1.2.5.3

Multiply $0$ by $0$ .

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

Step 7.3.1.3

Evaluate the limit of the denominator.

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Step 7.3.1.3.1

Evaluate the limit.

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Step 7.3.1.3.1.1

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

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

Step 7.3.1.3.1.2

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

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

Step 7.3.1.3.1.3

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

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

Step 7.3.1.3.2

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

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

Step 7.3.1.3.3

Simplify the answer.

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Step 7.3.1.3.3.1

Simplify each term.

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Step 7.3.1.3.3.1.1

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

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

Step 7.3.1.3.3.1.2

Multiply $- 1$ by $1$ .

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

Step 7.3.1.3.3.2

Subtract $1$ from $1$ .

$e^{- \frac{0}{0}}$

Step 7.3.1.3.3.3

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

Undefined

$e^{- \frac{0}{0}}$

Step 7.3.1.3.4

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

Undefined

$e^{- \frac{0}{0}}$

Step 7.3.1.4

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

Undefined

$e^{- \frac{0}{0}}$

Step 7.3.2

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.

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

Step 7.3.3

Find the derivative of the numerator and denominator.

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Step 7.3.3.1

Differentiate the numerator and denominator.

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

Step 7.3.3.2

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 $f (x) = x^{2}$ and $g (x) = \sin (x)$ .

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

Step 7.3.3.3

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

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

Step 7.3.3.4

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

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

Step 7.3.3.5

Reorder terms.

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

Step 7.3.3.6

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

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

Step 7.3.3.7

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

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

Step 7.3.3.8

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

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Step 7.3.3.8.1

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

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

Step 7.3.3.8.2

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

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

Step 7.3.3.8.3

Multiply $- 1$ by $- 1$ .

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

Step 7.3.3.8.4

Multiply $\sin (x)$ by $1$ .

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

Step 7.3.3.9

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

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

Step 7.4

Since $\frac{x^{2} \cdot - 1 + 2 x \sin (x)}{\sin (x)} \leq \frac{x^{2} \cos (x) + 2 x \sin (x)}{\sin (x)} \leq \frac{x^{2} \cdot 1 + 2 x \sin (x)}{\sin (x)}$ and $lim_{x \to 0^{+}} \frac{x^{2} \cdot - 1 + 2 x \sin (x)}{\sin (x)} = lim_{x \to 0^{+}} \frac{x^{2} \cdot 1 + 2 x \sin (x)}{\sin (x)} = 0$ , apply the squeeze theorem.

$e^{- 0}$

Step 7.5

Multiply $- 1$ by $0$ .

$e^{0}$

Step 8

Anything raised to $0$ is $1$ .

$1$