Calculus Examples

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

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.

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

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

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

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

Step 1.1.2.1.2

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

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

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

Step 1.1.2.3.2

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

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

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

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

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

Step 1.1.3.1.2

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

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

Step 1.1.3.1.3

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

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

Step 1.1.3.2

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

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

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

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

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

Step 1.1.3.3.1.2

Multiply $- 1$ by $1$ .

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

Step 1.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

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

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Step 1.3.2.1

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

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

Step 1.3.2.2

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

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

Step 1.3.2.3

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

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

Step 1.3.3

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

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

Step 1.3.4

Simplify.

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Step 1.3.4.1

Reorder the factors of $2 \sin (x) \cos (x)$ .

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

Step 1.3.4.2

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

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

Step 1.3.4.3

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

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

Step 1.3.4.4

Apply the sine double-angle identity.

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

Step 1.3.5

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

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

Step 1.3.6

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

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

Step 1.3.7

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

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Step 1.3.7.1

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

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

Step 1.3.7.2

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

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

Step 1.3.7.3

Multiply $- 1$ by $- 1$ .

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

Step 1.3.7.4

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

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

Step 1.3.8

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

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

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.

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

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.

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

Step 2.1.2.1.2

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Simplify the answer.

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Step 2.1.2.3.1

Multiply $2$ by $0$ .

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

Step 2.1.2.3.2

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

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

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

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

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

Step 2.1.3.2

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

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

Step 2.1.3.3

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

$\frac{0}{0}$

Step 2.1.3.4

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

Undefined

$\frac{0}{0}$

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

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 $2 x$ .

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $2$ by $1$ .

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

Step 2.3.6

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

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

Step 2.3.7

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

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

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 $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 4

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

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Step 4.1

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

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

Step 4.2

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

$2 \frac{\cos (2 \cdot 0)}{\cos (0)}$

Step 5

Simplify the answer.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Multiply $2$ by $0$ .

$2 \frac{\cos (0)}{\cos (0)}$

Step 5.1.2

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

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

Step 5.2

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

$2 (\frac{1}{1})$

Step 5.3

Cancel the common factor of $1$ .

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Step 5.3.1

Cancel the common factor.

$2 (\frac{1}{1})$

Step 5.3.2

Rewrite the expression.

$2 \cdot 1$

Step 5.4

Multiply $2$ by $1$ .

$2$