Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

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

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

Step 1.1.2.1.2

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

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

Step 1.1.2.1.3

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

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

Step 1.1.2.1.4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Multiply $2$ by $0$ .

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

Step 1.1.2.3.1.2

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

$\frac{1 - 1 \cdot 1}{lim_{x \to 0} x^{2}}$

Step 1.1.2.3.1.3

Multiply $- 1$ by $1$ .

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

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 0} x^{2}}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

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

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

Step 1.1.3.2

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

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

Step 1.1.3.3

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

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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Step 1.3.4.1

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

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

Step 1.3.4.2

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

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Step 1.3.4.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

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

Step 1.3.4.2.2

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

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

Step 1.3.4.2.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 1.3.4.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 1.3.4.4

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

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

Step 1.3.4.5

Multiply $2$ by $1$ .

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

Step 1.3.4.6

Multiply $2$ by $- 1$ .

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

Step 1.3.4.7

Multiply $- 2$ by $- 1$ .

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

Step 1.3.5

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

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

Step 1.3.6

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

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

Step 1.4

Cancel the common factor of $2$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Evaluate the limit.

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Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Simplify the answer.

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Step 2.1.2.3.1

Multiply $2$ by $0$ .

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

Step 2.1.2.3.2

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

$\frac{0}{lim_{x \to 0} x}$

Step 2.1.3

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

$\frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{0}{0}$

Step 2.2

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

Step 2.3.2

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

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Step 2.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

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

Step 2.3.2.2

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

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

Step 2.3.2.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 2.3.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $2$ by $1$ .

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

Step 2.3.6

Move $2$ to the left of $\cos (2 x)$ .

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

Step 2.3.7

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

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

Step 2.4

Divide $2 \cos (2 x)$ by $1$ .

$lim_{x \to 0} 2 \cos (2 x)$

Step 3

Evaluate the limit.

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Step 3.1

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 lim_{x \to 0} \cos (2 x)$

Step 3.2

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

$2 \cos (lim_{x \to 0} 2 x)$

Step 3.3

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \cos (2 lim_{x \to 0} x)$

Step 4

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

$2 \cos (2 \cdot 0)$

Step 5

Simplify the answer.

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Step 5.1

Multiply $2$ by $0$ .

$2 \cos (0)$

Step 5.2

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

$2 \cdot 1$

Step 5.3

Multiply $2$ by $1$ .

$2$