Calculus Examples

$lim_{x \to \infty} \frac{6 x^{4} - 5 x^{2}}{7 x^{4} + 14}$

Step 1

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

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Step 1.1

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

$\frac{lim_{x \to \infty} 6 x^{4} - 5 x^{2}}{lim_{x \to \infty} 7 x^{4} + 14}$

Step 1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\frac{\infty}{lim_{x \to \infty} 7 x^{4} + 14}$

Step 1.3

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\frac{\infty}{\infty}$

Step 1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 2

Since $\frac{\infty}{\infty}$ 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.

$lim_{x \to \infty} \frac{6 x^{4} - 5 x^{2}}{7 x^{4} + 14} = lim_{x \to \infty} \frac{\frac{d}{d x} [6 x^{4} - 5 x^{2}]}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [6 x^{4} - 5 x^{2}]}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3.2

By the Sum Rule, the derivative of $6 x^{4} - 5 x^{2}$ with respect to $x$ is $\frac{d}{d x} [6 x^{4}] + \frac{d}{d x} [- 5 x^{2}]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [6 x^{4}] + \frac{d}{d x} [- 5 x^{2}]}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3.3

Evaluate $\frac{d}{d x} [6 x^{4}]$ .

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Step 3.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{4}$ with respect to $x$ is $6 \frac{d}{d x} [x^{4}]$ .

$lim_{x \to \infty} \frac{6 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 5 x^{2}]}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3.3.2

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

$lim_{x \to \infty} \frac{6 (4 x^{3}) + \frac{d}{d x} [- 5 x^{2}]}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3.3.3

Multiply $4$ by $6$ .

$lim_{x \to \infty} \frac{24 x^{3} + \frac{d}{d x} [- 5 x^{2}]}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3.4

Evaluate $\frac{d}{d x} [- 5 x^{2}]$ .

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Step 3.4.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{2}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to \infty} \frac{24 x^{3} - 5 \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3.4.2

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

$lim_{x \to \infty} \frac{24 x^{3} - 5 (2 x)}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3.4.3

Multiply $2$ by $- 5$ .

$lim_{x \to \infty} \frac{24 x^{3} - 10 x}{\frac{d}{d x} [7 x^{4} + 14]}$

Step 3.5

By the Sum Rule, the derivative of $7 x^{4} + 14$ with respect to $x$ is $\frac{d}{d x} [7 x^{4}] + \frac{d}{d x} [14]$ .

$lim_{x \to \infty} \frac{24 x^{3} - 10 x}{\frac{d}{d x} [7 x^{4}] + \frac{d}{d x} [14]}$

Step 3.6

Evaluate $\frac{d}{d x} [7 x^{4}]$ .

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Step 3.6.1

Since $7$ is constant with respect to $x$ , the derivative of $7 x^{4}$ with respect to $x$ is $7 \frac{d}{d x} [x^{4}]$ .

$lim_{x \to \infty} \frac{24 x^{3} - 10 x}{7 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [14]}$

Step 3.6.2

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

$lim_{x \to \infty} \frac{24 x^{3} - 10 x}{7 (4 x^{3}) + \frac{d}{d x} [14]}$

Step 3.6.3

Multiply $4$ by $7$ .

$lim_{x \to \infty} \frac{24 x^{3} - 10 x}{28 x^{3} + \frac{d}{d x} [14]}$

Step 3.7

Since $14$ is constant with respect to $x$ , the derivative of $14$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{24 x^{3} - 10 x}{28 x^{3} + 0}$

Step 3.8

Add $28 x^{3}$ and $0$ .

$lim_{x \to \infty} \frac{24 x^{3} - 10 x}{28 x^{3}}$

Step 4

Move the term $\frac{1}{28}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{28} lim_{x \to \infty} \frac{24 x^{3} - 10 x}{x^{3}}$

Step 5

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^{3}$ .

$\frac{1}{28} lim_{x \to \infty} \frac{\frac{24 x^{3}}{x^{3}} + \frac{- 10 x}{x^{3}}}{\frac{x^{3}}{x^{3}}}$

Step 6

Evaluate the limit.

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Step 6.1

Simplify each term.

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Step 6.1.1

Cancel the common factor of $x^{3}$ .

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Step 6.1.1.1

Cancel the common factor.

$\frac{1}{28} lim_{x \to \infty} \frac{\frac{24 x^{3}}{x^{3}} + \frac{- 10 x}{x^{3}}}{\frac{x^{3}}{x^{3}}}$

Step 6.1.1.2

Divide $24$ by $1$ .

$\frac{1}{28} lim_{x \to \infty} \frac{24 + \frac{- 10 x}{x^{3}}}{\frac{x^{3}}{x^{3}}}$

Step 6.1.2

Cancel the common factor of $x$ and $x^{3}$ .

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Step 6.1.2.1

Factor $x$ out of $- 10 x$ .

$\frac{1}{28} lim_{x \to \infty} \frac{24 + \frac{x \cdot - 10}{x^{3}}}{\frac{x^{3}}{x^{3}}}$

Step 6.1.2.2

Cancel the common factors.

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Step 6.1.2.2.1

Factor $x$ out of $x^{3}$ .

$\frac{1}{28} lim_{x \to \infty} \frac{24 + \frac{x \cdot - 10}{x \cdot x^{2}}}{\frac{x^{3}}{x^{3}}}$

Step 6.1.2.2.2

Cancel the common factor.

$\frac{1}{28} lim_{x \to \infty} \frac{24 + \frac{x \cdot - 10}{x \cdot x^{2}}}{\frac{x^{3}}{x^{3}}}$

Step 6.1.2.2.3

Rewrite the expression.

$\frac{1}{28} lim_{x \to \infty} \frac{24 + \frac{- 10}{x^{2}}}{\frac{x^{3}}{x^{3}}}$

Step 6.1.3

Move the negative in front of the fraction.

$\frac{1}{28} lim_{x \to \infty} \frac{24 - \frac{10}{x^{2}}}{\frac{x^{3}}{x^{3}}}$

Step 6.2

Cancel the common factor of $x^{3}$ .

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Step 6.2.1

Cancel the common factor.

$\frac{1}{28} lim_{x \to \infty} \frac{24 - \frac{10}{x^{2}}}{\frac{x^{3}}{x^{3}}}$

Step 6.2.2

Rewrite the expression.

$\frac{1}{28} lim_{x \to \infty} \frac{24 - \frac{10}{x^{2}}}{1}$

Step 6.3

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

$\frac{1}{28} \cdot \frac{lim_{x \to \infty} 24 - \frac{10}{x^{2}}}{lim_{x \to \infty} 1}$

Step 6.4

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

$\frac{1}{28} \cdot \frac{lim_{x \to \infty} 24 - lim_{x \to \infty} \frac{10}{x^{2}}}{lim_{x \to \infty} 1}$

Step 6.5

Evaluate the limit of $24$ which is constant as $x$ approaches $\infty$ .

$\frac{1}{28} \cdot \frac{24 - lim_{x \to \infty} \frac{10}{x^{2}}}{lim_{x \to \infty} 1}$

Step 6.6

Move the term $10$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{28} \cdot \frac{24 - 10 lim_{x \to \infty} \frac{1}{x^{2}}}{lim_{x \to \infty} 1}$

Step 7

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{2}}$ approaches $0$ .

$\frac{1}{28} \cdot \frac{24 - 10 \cdot 0}{lim_{x \to \infty} 1}$

Step 8

Evaluate the limit.

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Step 8.1

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

$\frac{1}{28} \cdot \frac{24 - 10 \cdot 0}{1}$

Step 8.2

Simplify the answer.

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Step 8.2.1

Divide $24 - 10 \cdot 0$ by $1$ .

$\frac{1}{28} (24 - 10 \cdot 0)$

Step 8.2.2

Multiply $- 10$ by $0$ .

$\frac{1}{28} (24 + 0)$

Step 8.2.3

Add $24$ and $0$ .

$\frac{1}{28} \cdot 24$

Step 8.2.4

Cancel the common factor of $4$ .

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Step 8.2.4.1

Factor $4$ out of $28$ .

$\frac{1}{4 (7)} \cdot 24$

Step 8.2.4.2

Factor $4$ out of $24$ .

$\frac{1}{4 \cdot 7} \cdot (4 \cdot 6)$

Step 8.2.4.3

Cancel the common factor.

$\frac{1}{4 \cdot 7} \cdot (4 \cdot 6)$

Step 8.2.4.4

Rewrite the expression.

$\frac{1}{7} \cdot 6$

Step 8.2.5

Combine $\frac{1}{7}$ and $6$ .

$\frac{6}{7}$

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