Calculus Examples

$lim_{x \to 0} \frac{x^{2} - 5 x}{\sin (4 x)}$

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.

$\frac{lim_{x \to 0} x^{2} - 5 x}{lim_{x \to 0} \sin (4 x)}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

$\frac{lim_{x \to 0} x^{2} - lim_{x \to 0} 5 x}{lim_{x \to 0} \sin (4 x)}$

Step 1.1.2.2

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

$\frac{{(lim_{x \to 0} x)}^{2} - lim_{x \to 0} 5 x}{lim_{x \to 0} \sin (4 x)}$

Step 1.1.2.3

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

$\frac{{(lim_{x \to 0} x)}^{2} - 5 lim_{x \to 0} x}{lim_{x \to 0} \sin (4 x)}$

Step 1.1.2.4

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

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Step 1.1.2.4.1

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

$\frac{0^{2} - 5 lim_{x \to 0} x}{lim_{x \to 0} \sin (4 x)}$

Step 1.1.2.4.2

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

$\frac{0^{2} - 5 \cdot 0}{lim_{x \to 0} \sin (4 x)}$

Step 1.1.2.5

Simplify the answer.

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Step 1.1.2.5.1

Simplify each term.

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Step 1.1.2.5.1.1

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

$\frac{0 - 5 \cdot 0}{lim_{x \to 0} \sin (4 x)}$

Step 1.1.2.5.1.2

Multiply $- 5$ by $0$ .

$\frac{0 + 0}{lim_{x \to 0} \sin (4 x)}$

Step 1.1.2.5.2

Add $0$ and $0$ .

$\frac{0}{lim_{x \to 0} \sin (4 x)}$

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

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

$\frac{0}{\sin (lim_{x \to 0} 4 x)}$

Step 1.1.3.1.2

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{\sin (4 lim_{x \to 0} x)}$

Step 1.1.3.2

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

$\frac{0}{\sin (4 \cdot 0)}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Multiply $4$ by $0$ .

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

Step 1.1.3.3.2

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

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{x^{2} - 5 x}{\sin (4 x)} = lim_{x \to 0} \frac{\frac{d}{d x} [x^{2} - 5 x]}{\frac{d}{d x} [\sin (4 x)]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [x^{2} - 5 x]}{\frac{d}{d x} [\sin (4 x)]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} - 5 x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 5 x]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [\sin (4 x)]}$

Step 1.3.3

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

$lim_{x \to 0} \frac{2 x + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [\sin (4 x)]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- 5 x]$ .

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Step 1.3.4.1

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

$lim_{x \to 0} \frac{2 x - 5 \frac{d}{d x} [x]}{\frac{d}{d x} [\sin (4 x)]}$

Step 1.3.4.2

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

$lim_{x \to 0} \frac{2 x - 5 \cdot 1}{\frac{d}{d x} [\sin (4 x)]}$

Step 1.3.4.3

Multiply $- 5$ by $1$ .

$lim_{x \to 0} \frac{2 x - 5}{\frac{d}{d x} [\sin (4 x)]}$

Step 1.3.5

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 $g (x) = 4 x$ .

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Step 1.3.5.1

To apply the Chain Rule, set $u$ as $4 x$ .

$lim_{x \to 0} \frac{2 x - 5}{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [4 x]}$

Step 1.3.5.2

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

$lim_{x \to 0} \frac{2 x - 5}{\cos (u) \frac{d}{d x} [4 x]}$

Step 1.3.5.3

Replace all occurrences of $u$ with $4 x$ .

$lim_{x \to 0} \frac{2 x - 5}{\cos (4 x) \frac{d}{d x} [4 x]}$

Step 1.3.6

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{2 x - 5}{\cos (4 x) (4 \frac{d}{d x} [x])}$

Step 1.3.7

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

$lim_{x \to 0} \frac{2 x - 5}{\cos (4 x) (4 \cdot 1)}$

Step 1.3.8

Multiply $4$ by $1$ .

$lim_{x \to 0} \frac{2 x - 5}{\cos (4 x) \cdot 4}$

Step 1.3.9

Move $4$ to the left of $\cos (4 x)$ .

$lim_{x \to 0} \frac{2 x - 5}{4 \cos (4 x)}$

Step 2

Evaluate the limit.

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Step 2.1

Move the term $\frac{1}{4}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{4} lim_{x \to 0} \frac{2 x - 5}{\cos (4 x)}$

Step 2.2

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

$\frac{1}{4} \cdot \frac{lim_{x \to 0} 2 x - 5}{lim_{x \to 0} \cos (4 x)}$

Step 2.3

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

$\frac{1}{4} \cdot \frac{lim_{x \to 0} 2 x - lim_{x \to 0} 5}{lim_{x \to 0} \cos (4 x)}$

Step 2.4

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

$\frac{1}{4} \cdot \frac{2 lim_{x \to 0} x - lim_{x \to 0} 5}{lim_{x \to 0} \cos (4 x)}$

Step 2.5

Evaluate the limit of $5$ which is constant as $x$ approaches $0$ .

$\frac{1}{4} \cdot \frac{2 lim_{x \to 0} x - 1 \cdot 5}{lim_{x \to 0} \cos (4 x)}$

Step 2.6

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

$\frac{1}{4} \cdot \frac{2 lim_{x \to 0} x - 1 \cdot 5}{\cos (lim_{x \to 0} 4 x)}$

Step 2.7

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{4} \cdot \frac{2 lim_{x \to 0} x - 1 \cdot 5}{\cos (4 lim_{x \to 0} x)}$

Step 3

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

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Step 3.1

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

$\frac{1}{4} \cdot \frac{2 \cdot 0 - 1 \cdot 5}{\cos (4 lim_{x \to 0} x)}$

Step 3.2

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

$\frac{1}{4} \cdot \frac{2 \cdot 0 - 1 \cdot 5}{\cos (4 \cdot 0)}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $2$ by $0$ .

$\frac{1}{4} \cdot \frac{0 - 1 \cdot 5}{\cos (4 \cdot 0)}$

Step 4.1.2

Multiply $- 1$ by $5$ .

$\frac{1}{4} \cdot \frac{0 - 5}{\cos (4 \cdot 0)}$

Step 4.1.3

Subtract $5$ from $0$ .

$\frac{1}{4} \cdot \frac{- 5}{\cos (4 \cdot 0)}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $4$ by $0$ .

$\frac{1}{4} \cdot \frac{- 5}{\cos (0)}$

Step 4.2.2

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

$\frac{1}{4} \cdot \frac{- 5}{1}$

Step 4.3

Divide $- 5$ by $1$ .

$\frac{1}{4} \cdot - 5$

Step 4.4

Combine $\frac{1}{4}$ and $- 5$ .

$\frac{- 5}{4}$

Step 4.5

Move the negative in front of the fraction.

$- \frac{5}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$- \frac{5}{4}$

Decimal Form:

$- 1.25$