Calculus Examples

$lim_{x \to 0} \frac{\sin (5 x)}{\sin (2 x)}$

Step 1

Multiply the numerator and denominator by $2 x$ .

$lim_{x \to 0} \frac{\sin (5 x) \cdot (2 x)}{\sin (2 x) \cdot (2 x)}$

Step 2

Multiply the numerator and denominator by $5 x$ .

$lim_{x \to 0} \frac{\sin (5 x) \cdot (2 x) \cdot (5 x)}{5 x \cdot \sin (2 x) \cdot (2 x)}$

Step 3

Separate fractions.

$lim_{x \to 0} \frac{\sin (5 x)}{5 x} \cdot \frac{2 x}{\sin (2 x)} \cdot \frac{5 x}{2 x}$

Step 4

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

$lim_{x \to 0} \frac{\sin (5 x)}{5 x} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5

The limit of $\frac{\sin (5 x)}{5 x}$ as $x$ approaches $0$ is $1$ .

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Step 5.1

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

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Step 5.1.1

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

$\frac{lim_{x \to 0} \sin (5 x)}{lim_{x \to 0} 5 x} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.2

Evaluate the limit of the numerator.

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Step 5.1.2.1

Evaluate the limit.

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Step 5.1.2.1.1

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

$\frac{\sin (lim_{x \to 0} 5 x)}{lim_{x \to 0} 5 x} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.2.1.2

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

$\frac{\sin (5 lim_{x \to 0} x)}{lim_{x \to 0} 5 x} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.2.2

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

$\frac{\sin (5 \cdot 0)}{lim_{x \to 0} 5 x} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.2.3

Simplify the answer.

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Step 5.1.2.3.1

Multiply $5$ by $0$ .

$\frac{\sin (0)}{lim_{x \to 0} 5 x} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.2.3.2

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

$\frac{0}{lim_{x \to 0} 5 x} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.3

Evaluate the limit of the denominator.

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Step 5.1.3.1

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

$\frac{0}{5 lim_{x \to 0} x} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.3.2

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

$\frac{0}{5 \cdot 0} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.3.3

Multiply $5$ by $0$ .

$\frac{0}{0} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.3.4

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

Undefined

$\frac{0}{0} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.1.4

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

Undefined

$\frac{0}{0} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.2

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.

$lim_{x \to 0} \frac{\sin (5 x)}{5 x} = lim_{x \to 0} \frac{\frac{d}{d x} [\sin (5 x)]}{\frac{d}{d x} [5 x]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [\sin (5 x)]}{\frac{d}{d x} [5 x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.2

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 $g (x) = 5 x$ .

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Step 5.3.2.1

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

$lim_{x \to 0} \frac{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [5 x]}{\frac{d}{d x} [5 x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.2.2

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

$lim_{x \to 0} \frac{\cos (u) \frac{d}{d x} [5 x]}{\frac{d}{d x} [5 x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.2.3

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

$lim_{x \to 0} \frac{\cos (5 x) \frac{d}{d x} [5 x]}{\frac{d}{d x} [5 x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.3

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

$lim_{x \to 0} \frac{\cos (5 x) (5 \frac{d}{d x} [x])}{\frac{d}{d x} [5 x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.4

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

$lim_{x \to 0} \frac{\cos (5 x) (5 \cdot 1)}{\frac{d}{d x} [5 x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.5

Multiply $5$ by $1$ .

$lim_{x \to 0} \frac{\cos (5 x) \cdot 5}{\frac{d}{d x} [5 x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.6

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

$lim_{x \to 0} \frac{5 \cdot \cos (5 x)}{\frac{d}{d x} [5 x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.7

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

$lim_{x \to 0} \frac{5 \cos (5 x)}{5 \frac{d}{d x} [x]} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.8

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

$lim_{x \to 0} \frac{5 \cos (5 x)}{5 \cdot 1} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.3.9

Multiply $5$ by $1$ .

$lim_{x \to 0} \frac{5 \cos (5 x)}{5} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.4

Evaluate the limit.

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Step 5.4.1

Cancel the common factor of $5$ .

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Step 5.4.1.1

Cancel the common factor.

$lim_{x \to 0} \frac{5 \cos (5 x)}{5} \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.4.1.2

Divide $\cos (5 x)$ by $1$ .

$lim_{x \to 0} \cos (5 x) \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.4.2

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

$\cos (lim_{x \to 0} 5 x) \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.4.3

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

$\cos (5 lim_{x \to 0} x) \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.5

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

$\cos (5 \cdot 0) \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.6

Simplify the answer.

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Step 5.6.1

Multiply $5$ by $0$ .

$\cos (0) \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 5.6.2

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

$1 \cdot lim_{x \to 0} \frac{2 x}{\sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6

The limit of $\frac{2 x}{\sin (2 x)}$ as $x$ approaches $0$ is $1$ .

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Step 6.1

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

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Step 6.1.1

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

$1 \cdot \frac{lim_{x \to 0} 2 x}{lim_{x \to 0} \sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.2

Evaluate the limit of the numerator.

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Step 6.1.2.1

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

$1 \cdot \frac{2 lim_{x \to 0} x}{lim_{x \to 0} \sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.2.2

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

$1 \cdot \frac{2 \cdot 0}{lim_{x \to 0} \sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.2.3

Multiply $2$ by $0$ .

$1 \cdot \frac{0}{lim_{x \to 0} \sin (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.3

Evaluate the limit of the denominator.

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Step 6.1.3.1

Evaluate the limit.

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Step 6.1.3.1.1

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

$1 \cdot \frac{0}{\sin (lim_{x \to 0} 2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.3.1.2

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

$1 \cdot \frac{0}{\sin (2 lim_{x \to 0} x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.3.2

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

$1 \cdot \frac{0}{\sin (2 \cdot 0)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.3.3

Simplify the answer.

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Step 6.1.3.3.1

Multiply $2$ by $0$ .

$1 \cdot \frac{0}{\sin (0)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.3.3.2

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

$1 \cdot \frac{0}{0} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.3.3.3

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

Undefined

$1 \cdot \frac{0}{0} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.3.4

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

Undefined

$1 \cdot \frac{0}{0} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.1.4

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

Undefined

$1 \cdot \frac{0}{0} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.2

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.

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

Step 6.3

Find the derivative of the numerator and denominator.

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Step 6.3.1

Differentiate the numerator and denominator.

$1 \cdot lim_{x \to 0} \frac{\frac{d}{d x} [2 x]}{\frac{d}{d x} [\sin (2 x)]} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.2

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

$1 \cdot lim_{x \to 0} \frac{2 \frac{d}{d x} [x]}{\frac{d}{d x} [\sin (2 x)]} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.3

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

$1 \cdot lim_{x \to 0} \frac{2 \cdot 1}{\frac{d}{d x} [\sin (2 x)]} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.4

Multiply $2$ by $1$ .

$1 \cdot lim_{x \to 0} \frac{2}{\frac{d}{d x} [\sin (2 x)]} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.5

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 6.3.5.1

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

$1 \cdot lim_{x \to 0} \frac{2}{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x]} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.5.2

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

$1 \cdot lim_{x \to 0} \frac{2}{\cos (u) \frac{d}{d x} [2 x]} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.5.3

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

$1 \cdot lim_{x \to 0} \frac{2}{\cos (2 x) \frac{d}{d x} [2 x]} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.6

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

$1 \cdot lim_{x \to 0} \frac{2}{\cos (2 x) (2 \frac{d}{d x} [x])} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.7

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

$1 \cdot lim_{x \to 0} \frac{2}{\cos (2 x) (2 \cdot 1)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.8

Multiply $2$ by $1$ .

$1 \cdot lim_{x \to 0} \frac{2}{\cos (2 x) \cdot 2} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.3.9

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

$1 \cdot lim_{x \to 0} \frac{2}{2 \cos (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.4

Cancel the common factor of $2$ .

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Step 6.4.1

Cancel the common factor.

$1 \cdot lim_{x \to 0} \frac{2}{2 \cos (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.4.2

Rewrite the expression.

$1 \cdot lim_{x \to 0} \frac{1}{\cos (2 x)} \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.5

Convert from $\frac{1}{\cos (2 x)}$ to $\sec (2 x)$ .

$1 \cdot lim_{x \to 0} \sec (2 x) \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.6

Evaluate the limit.

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Step 6.6.1

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

$1 \cdot \sec (lim_{x \to 0} 2 x) \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.6.2

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

$1 \cdot \sec (2 lim_{x \to 0} x) \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.7

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

$1 \cdot \sec (2 \cdot 0) \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.8

Simplify the answer.

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Step 6.8.1

Multiply $2$ by $0$ .

$1 \cdot \sec (0) \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 6.8.2

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

$1 \cdot 1 \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 7

Cancel the common factor of $x$ .

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Step 7.1

Cancel the common factor.

$1 \cdot 1 \cdot lim_{x \to 0} \frac{5 x}{2 x}$

Step 7.2

Rewrite the expression.

$1 \cdot 1 \cdot lim_{x \to 0} \frac{5}{2}$

Step 8

Evaluate the limit of $\frac{5}{2}$ which is constant as $x$ approaches $0$ .

$1 \cdot 1 \cdot \frac{5}{2}$

Step 9

Simplify the answer.

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Step 9.1

Multiply $1$ by $1$ .

$1 \cdot \frac{5}{2}$

Step 9.2

Multiply $\frac{5}{2}$ by $1$ .

$\frac{5}{2}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$\frac{5}{2}$

Decimal Form:

$2.5$