Calculus Examples

$lim_{x \to 0} \frac{9 x^{2} + 4 x^{5}}{- 7 x^{2}}$

Step 1

Move the term $\frac{1}{- 7}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 7} lim_{x \to 0} \frac{9 x^{2} + 4 x^{5}}{x^{2}}$

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.

$\frac{1}{- 7} \cdot \frac{lim_{x \to 0} 9 x^{2} + 4 x^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

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

$\frac{1}{- 7} \cdot \frac{lim_{x \to 0} 9 x^{2} + lim_{x \to 0} 4 x^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.2

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 7} \cdot \frac{9 lim_{x \to 0} x^{2} + lim_{x \to 0} 4 x^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.3

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

$\frac{1}{- 7} \cdot \frac{9 {(lim_{x \to 0} x)}^{2} + lim_{x \to 0} 4 x^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.4

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

$\frac{1}{- 7} \cdot \frac{9 {(lim_{x \to 0} x)}^{2} + 4 lim_{x \to 0} x^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.5

Move the exponent $5$ from $x^{5}$ outside the limit using the Limits Power Rule.

$\frac{1}{- 7} \cdot \frac{9 {(lim_{x \to 0} x)}^{2} + 4 {(lim_{x \to 0} x)}^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.6

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

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Step 2.1.2.6.1

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

$\frac{1}{- 7} \cdot \frac{9 \cdot 0^{2} + 4 {(lim_{x \to 0} x)}^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.6.2

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

$\frac{1}{- 7} \cdot \frac{9 \cdot 0^{2} + 4 \cdot 0^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.7

Simplify the answer.

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Step 2.1.2.7.1

Simplify each term.

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Step 2.1.2.7.1.1

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

$\frac{1}{- 7} \cdot \frac{9 \cdot 0 + 4 \cdot 0^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.7.1.2

Multiply $9$ by $0$ .

$\frac{1}{- 7} \cdot \frac{0 + 4 \cdot 0^{5}}{lim_{x \to 0} x^{2}}$

Step 2.1.2.7.1.3

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

$\frac{1}{- 7} \cdot \frac{0 + 4 \cdot 0}{lim_{x \to 0} x^{2}}$

Step 2.1.2.7.1.4

Multiply $4$ by $0$ .

$\frac{1}{- 7} \cdot \frac{0 + 0}{lim_{x \to 0} x^{2}}$

Step 2.1.2.7.2

Add $0$ and $0$ .

$\frac{1}{- 7} \cdot \frac{0}{lim_{x \to 0} x^{2}}$

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

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

$\frac{1}{- 7} \cdot \frac{0}{{(lim_{x \to 0} x)}^{2}}$

Step 2.1.3.2

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

$\frac{1}{- 7} \cdot \frac{0}{0^{2}}$

Step 2.1.3.3

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

$\frac{1}{- 7} \cdot \frac{0}{0}$

Step 2.1.3.4

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

Undefined

$\frac{1}{- 7} \cdot \frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{1}{- 7} \cdot \frac{0}{0}$

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.

$lim_{x \to 0} \frac{9 x^{2} + 4 x^{5}}{x^{2}} = lim_{x \to 0} \frac{\frac{d}{d x} [9 x^{2} + 4 x^{5}]}{\frac{d}{d x} [x^{2}]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$\frac{1}{- 7} lim_{x \to 0} \frac{\frac{d}{d x} [9 x^{2} + 4 x^{5}]}{\frac{d}{d x} [x^{2}]}$

Step 2.3.2

By the Sum Rule, the derivative of $9 x^{2} + 4 x^{5}$ with respect to $x$ is $\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [4 x^{5}]$ .

$\frac{1}{- 7} lim_{x \to 0} \frac{\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [4 x^{5}]}{\frac{d}{d x} [x^{2}]}$

Step 2.3.3

Evaluate $\frac{d}{d x} [9 x^{2}]$ .

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Step 2.3.3.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

$\frac{1}{- 7} lim_{x \to 0} \frac{9 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x^{5}]}{\frac{d}{d x} [x^{2}]}$

Step 2.3.3.2

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

$\frac{1}{- 7} lim_{x \to 0} \frac{9 \cdot 2 x + \frac{d}{d x} [4 x^{5}]}{\frac{d}{d x} [x^{2}]}$

Step 2.3.3.3

Multiply $2$ by $9$ .

$\frac{1}{- 7} lim_{x \to 0} \frac{18 x + \frac{d}{d x} [4 x^{5}]}{\frac{d}{d x} [x^{2}]}$

Step 2.3.4

Evaluate $\frac{d}{d x} [4 x^{5}]$ .

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Step 2.3.4.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{5}$ with respect to $x$ is $4 \frac{d}{d x} [x^{5}]$ .

$\frac{1}{- 7} lim_{x \to 0} \frac{18 x + 4 \frac{d}{d x} [x^{5}]}{\frac{d}{d x} [x^{2}]}$

Step 2.3.4.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$\frac{1}{- 7} lim_{x \to 0} \frac{18 x + 4 \cdot 5 x^{4}}{\frac{d}{d x} [x^{2}]}$

Step 2.3.4.3

Multiply $5$ by $4$ .

$\frac{1}{- 7} lim_{x \to 0} \frac{18 x + 20 x^{4}}{\frac{d}{d x} [x^{2}]}$

Step 2.3.5

Reorder terms.

$\frac{1}{- 7} lim_{x \to 0} \frac{20 x^{4} + 18 x}{\frac{d}{d x} [x^{2}]}$

Step 2.3.6

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

$\frac{1}{- 7} lim_{x \to 0} \frac{20 x^{4} + 18 x}{2 x}$

Step 3

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{20 x^{4} + 18 x}{x}$

Step 4

Apply L'Hospital's rule.

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Step 4.1

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

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Step 4.1.1

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

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{lim_{x \to 0} 20 x^{4} + 18 x}{lim_{x \to 0} x}$

Step 4.1.2

Evaluate the limit of the numerator.

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Step 4.1.2.1

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

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{lim_{x \to 0} 20 x^{4} + lim_{x \to 0} 18 x}{lim_{x \to 0} x}$

Step 4.1.2.2

Move the term $20$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{20 lim_{x \to 0} x^{4} + lim_{x \to 0} 18 x}{lim_{x \to 0} x}$

Step 4.1.2.3

Move the exponent $4$ from $x^{4}$ outside the limit using the Limits Power Rule.

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{20 {(lim_{x \to 0} x)}^{4} + lim_{x \to 0} 18 x}{lim_{x \to 0} x}$

Step 4.1.2.4

Move the term $18$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{20 {(lim_{x \to 0} x)}^{4} + 18 lim_{x \to 0} x}{lim_{x \to 0} x}$

Step 4.1.2.5

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

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Step 4.1.2.5.1

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

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{20 \cdot 0^{4} + 18 lim_{x \to 0} x}{lim_{x \to 0} x}$

Step 4.1.2.5.2

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

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{20 \cdot 0^{4} + 18 \cdot 0}{lim_{x \to 0} x}$

Step 4.1.2.6

Simplify the answer.

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Step 4.1.2.6.1

Simplify each term.

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Step 4.1.2.6.1.1

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

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{20 \cdot 0 + 18 \cdot 0}{lim_{x \to 0} x}$

Step 4.1.2.6.1.2

Multiply $20$ by $0$ .

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{0 + 18 \cdot 0}{lim_{x \to 0} x}$

Step 4.1.2.6.1.3

Multiply $18$ by $0$ .

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{0 + 0}{lim_{x \to 0} x}$

Step 4.1.2.6.2

Add $0$ and $0$ .

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{0}{lim_{x \to 0} x}$

Step 4.1.3

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

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{0}{0}$

Step 4.1.4

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

Undefined

$\frac{1}{- 7} \cdot \frac{1}{2} \frac{0}{0}$

Step 4.2

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.

$lim_{x \to 0} \frac{20 x^{4} + 18 x}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [20 x^{4} + 18 x]}{\frac{d}{d x} [x]}$

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{\frac{d}{d x} [20 x^{4} + 18 x]}{\frac{d}{d x} [x]}$

Step 4.3.2

By the Sum Rule, the derivative of $20 x^{4} + 18 x$ with respect to $x$ is $\frac{d}{d x} [20 x^{4}] + \frac{d}{d x} [18 x]$ .

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{\frac{d}{d x} [20 x^{4}] + \frac{d}{d x} [18 x]}{\frac{d}{d x} [x]}$

Step 4.3.3

Evaluate $\frac{d}{d x} [20 x^{4}]$ .

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Step 4.3.3.1

Since $20$ is constant with respect to $x$ , the derivative of $20 x^{4}$ with respect to $x$ is $20 \frac{d}{d x} [x^{4}]$ .

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{20 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [18 x]}{\frac{d}{d x} [x]}$

Step 4.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{20 \cdot 4 x^{3} + \frac{d}{d x} [18 x]}{\frac{d}{d x} [x]}$

Step 4.3.3.3

Multiply $4$ by $20$ .

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{80 x^{3} + \frac{d}{d x} [18 x]}{\frac{d}{d x} [x]}$

Step 4.3.4

Evaluate $\frac{d}{d x} [18 x]$ .

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Step 4.3.4.1

Since $18$ is constant with respect to $x$ , the derivative of $18 x$ with respect to $x$ is $18 \frac{d}{d x} [x]$ .

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{80 x^{3} + 18 \frac{d}{d x} [x]}{\frac{d}{d x} [x]}$

Step 4.3.4.2

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

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{80 x^{3} + 18 \cdot 1}{\frac{d}{d x} [x]}$

Step 4.3.4.3

Multiply $18$ by $1$ .

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{80 x^{3} + 18}{\frac{d}{d x} [x]}$

Step 4.3.5

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

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} \frac{80 x^{3} + 18}{1}$

Step 4.4

Divide $80 x^{3} + 18$ by $1$ .

$\frac{1}{- 7} \cdot \frac{1}{2} lim_{x \to 0} 80 x^{3} + 18$

Step 5

Evaluate the limit.

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Step 5.1

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

$\frac{1}{- 7} \cdot \frac{1}{2} (lim_{x \to 0} 80 x^{3} + lim_{x \to 0} 18)$

Step 5.2

Move the term $80$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 7} \cdot \frac{1}{2} (80 lim_{x \to 0} x^{3} + lim_{x \to 0} 18)$

Step 5.3

Move the exponent $3$ from $x^{3}$ outside the limit using the Limits Power Rule.

$\frac{1}{- 7} \cdot \frac{1}{2} (80 {(lim_{x \to 0} x)}^{3} + lim_{x \to 0} 18)$

Step 5.4

Evaluate the limit of $18$ which is constant as $x$ approaches $0$ .

$\frac{1}{- 7} \cdot \frac{1}{2} (80 {(lim_{x \to 0} x)}^{3} + 18)$

Step 6

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

$\frac{1}{- 7} \cdot \frac{1}{2} (80 \cdot 0^{3} + 18)$

Step 7

Simplify the answer.

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Step 7.1

Move the negative in front of the fraction.

$- \frac{1}{7} \cdot \frac{1}{2} (80 \cdot 0^{3} + 18)$

Step 7.2

Multiply $- \frac{1}{7} \cdot \frac{1}{2}$ .

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Step 7.2.1

Multiply $\frac{1}{2}$ by $\frac{1}{7}$ .

$- \frac{1}{2 \cdot 7} (80 \cdot 0^{3} + 18)$

Step 7.2.2

Multiply $2$ by $7$ .

$- \frac{1}{14} (80 \cdot 0^{3} + 18)$

Step 7.3

Simplify each term.

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Step 7.3.1

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

$- \frac{1}{14} (80 \cdot 0 + 18)$

Step 7.3.2

Multiply $80$ by $0$ .

$- \frac{1}{14} (0 + 18)$

Step 7.4

Add $0$ and $18$ .

$- \frac{1}{14} \cdot 18$

Step 7.5

Cancel the common factor of $2$ .

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Step 7.5.1

Move the leading negative in $- \frac{1}{14}$ into the numerator.

$\frac{- 1}{14} \cdot 18$

Step 7.5.2

Factor $2$ out of $14$ .

$\frac{- 1}{2 (7)} \cdot 18$

Step 7.5.3

Factor $2$ out of $18$ .

$\frac{- 1}{2 \cdot 7} \cdot (2 \cdot 9)$

Step 7.5.4

Cancel the common factor.

$\frac{- 1}{2 \cdot 7} \cdot (2 \cdot 9)$

Step 7.5.5

Rewrite the expression.

$\frac{- 1}{7} \cdot 9$

Step 7.6

Combine $\frac{- 1}{7}$ and $9$ .

$\frac{- 1 \cdot 9}{7}$

Step 7.7

Multiply $- 1$ by $9$ .

$\frac{- 9}{7}$

Step 7.8

Move the negative in front of the fraction.

$- \frac{9}{7}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$- \frac{9}{7}$

Decimal Form:

$- 1.\overline{285714}$