Calculus Examples

$lim_{x \to - \infty} \frac{3 x}{\sqrt{16 x^{2} - 9 x}}$

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.

$\frac{lim_{x \to - \infty} 3 x}{lim_{x \to - \infty} \sqrt{16 x^{2} - 9 x}}$

Step 1.2

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

$\frac{- \infty}{lim_{x \to - \infty} \sqrt{16 x^{2} - 9 x}}$

Step 1.3

As $x$ approaches $- \infty$ for radicals, the value goes to $\infty$ .

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

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.

$lim_{x \to - \infty} \frac{3 x}{\sqrt{16 x^{2} - 9 x}} = lim_{x \to - \infty} \frac{\frac{d}{d x} [3 x]}{\frac{d}{d x} [\sqrt{16 x^{2} - 9 x}]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to - \infty} \frac{\frac{d}{d x} [3 x]}{\frac{d}{d x} [\sqrt{16 x^{2} - 9 x}]}$

Step 3.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$lim_{x \to - \infty} \frac{3 \frac{d}{d x} [x]}{\frac{d}{d x} [\sqrt{16 x^{2} - 9 x}]}$

Step 3.3

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

$lim_{x \to - \infty} \frac{3 \cdot 1}{\frac{d}{d x} [\sqrt{16 x^{2} - 9 x}]}$

Step 3.4

Multiply $3$ by $1$ .

$lim_{x \to - \infty} \frac{3}{\frac{d}{d x} [\sqrt{16 x^{2} - 9 x}]}$

Step 3.5

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{16 x^{2} - 9 x}$ as ${(16 x^{2} - 9 x)}^{\frac{1}{2}}$ .

$lim_{x \to - \infty} \frac{3}{\frac{d}{d x} [{(16 x^{2} - 9 x)}^{\frac{1}{2}}]}$

Step 3.6

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = 16 x^{2} - 9 x$ .

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Step 3.6.1

To apply the Chain Rule, set $u$ as $16 x^{2} - 9 x$ .

$lim_{x \to - \infty} \frac{3}{\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.6.2

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

$lim_{x \to - \infty} \frac{3}{\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.6.3

Replace all occurrences of $u$ with $16 x^{2} - 9 x$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2} {(16 x^{2} - 9 x)}^{\frac{1}{2} - 1} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2} {(16 x^{2} - 9 x)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.8

Combine $- 1$ and $\frac{2}{2}$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2} {(16 x^{2} - 9 x)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.9

Combine the numerators over the common denominator.

$lim_{x \to - \infty} \frac{3}{\frac{1}{2} {(16 x^{2} - 9 x)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.10

Simplify the numerator.

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Step 3.10.1

Multiply $- 1$ by $2$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2} {(16 x^{2} - 9 x)}^{\frac{1 - 2}{2}} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.10.2

Subtract $2$ from $1$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2} {(16 x^{2} - 9 x)}^{\frac{- 1}{2}} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.11

Move the negative in front of the fraction.

$lim_{x \to - \infty} \frac{3}{\frac{1}{2} {(16 x^{2} - 9 x)}^{- \frac{1}{2}} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.12

Combine $\frac{1}{2}$ and ${(16 x^{2} - 9 x)}^{- \frac{1}{2}}$ .

$lim_{x \to - \infty} \frac{3}{\frac{{(16 x^{2} - 9 x)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.13

Move ${(16 x^{2} - 9 x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} \frac{d}{d x} [16 x^{2} - 9 x]}$

Step 3.14

By the Sum Rule, the derivative of $16 x^{2} - 9 x$ with respect to $x$ is $\frac{d}{d x} [16 x^{2}] + \frac{d}{d x} [- 9 x]$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} (\frac{d}{d x} [16 x^{2}] + \frac{d}{d x} [- 9 x])}$

Step 3.15

Since $16$ is constant with respect to $x$ , the derivative of $16 x^{2}$ with respect to $x$ is $16 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} (16 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x])}$

Step 3.16

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

$lim_{x \to - \infty} \frac{3}{\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} (16 (2 x) + \frac{d}{d x} [- 9 x])}$

Step 3.17

Multiply $2$ by $16$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} (32 x + \frac{d}{d x} [- 9 x])}$

Step 3.18

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

$lim_{x \to - \infty} \frac{3}{\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} (32 x - 9 \frac{d}{d x} [x])}$

Step 3.19

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

$lim_{x \to - \infty} \frac{3}{\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} (32 x - 9 \cdot 1)}$

Step 3.20

Multiply $- 9$ by $1$ .

$lim_{x \to - \infty} \frac{3}{\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} (32 x - 9)}$

Step 3.21

Simplify.

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Step 3.21.1

Reorder the factors of $\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}} (32 x - 9)$ .

$lim_{x \to - \infty} \frac{3}{(32 x - 9) \frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}}}$

Step 3.21.2

Multiply $32 x - 9$ by $\frac{1}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}}$ .

$lim_{x \to - \infty} \frac{3}{\frac{32 x - 9}{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to - \infty} 3 \frac{2 {(16 x^{2} - 9 x)}^{\frac{1}{2}}}{32 x - 9}$

Step 5

Rewrite ${(16 x^{2} - 9 x)}^{\frac{1}{2}}$ as $\sqrt{16 x^{2} - 9 x}$ .

$lim_{x \to - \infty} 3 \frac{2 \sqrt{16 x^{2} - 9 x}}{32 x - 9}$

Step 6

Combine factors.

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Step 6.1

Combine $3$ and $\frac{2 \sqrt{16 x^{2} - 9 x}}{32 x - 9}$ .

$lim_{x \to - \infty} \frac{3 (2 \sqrt{16 x^{2} - 9 x})}{32 x - 9}$

Step 6.2

Multiply $2$ by $3$ .

$lim_{x \to - \infty} \frac{6 \sqrt{16 x^{2} - 9 x}}{32 x - 9}$

Step 7

Evaluate the limit.

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Step 7.1

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$6 lim_{x \to - \infty} \frac{\sqrt{16 x^{2} - 9 x}}{32 x - 9}$

Step 7.2

Factor $x$ out of $16 x^{2} - 9 x$ .

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Step 7.2.1

Factor $x$ out of $16 x^{2}$ .

$6 lim_{x \to - \infty} \frac{\sqrt{x \cdot 16 x - 9 x}}{32 x - 9}$

Step 7.2.2

Factor $x$ out of $- 9 x$ .

$6 lim_{x \to - \infty} \frac{\sqrt{x \cdot 16 x + x \cdot - 9}}{32 x - 9}$

Step 7.2.3

Factor $x$ out of $x (16 x) + x \cdot - 9$ .

$6 lim_{x \to - \infty} \frac{\sqrt{x (16 x - 9)}}{32 x - 9}$

Step 8

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $- \sqrt{x^{2}} = x$ .

$6 lim_{x \to - \infty} \frac{- \sqrt{\frac{x (16 x - 9)}{x^{2}}}}{\frac{32 x}{x} + \frac{- 9}{x}}$

Step 9

Simplify each term.

$6 lim_{x \to - \infty} \frac{- \sqrt{\frac{x (16 x - 9)}{x^{2}}}}{32 - \frac{9}{x}}$

Step 10

Cancel the common factors.

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Step 10.1

Factor $x$ out of $x^{2}$ .

$6 lim_{x \to - \infty} \frac{- \sqrt{\frac{x (16 x - 9)}{x \cdot x}}}{32 - \frac{9}{x}}$

Step 10.2

Cancel the common factor.

$6 lim_{x \to - \infty} \frac{- \sqrt{\frac{x (16 x - 9)}{x \cdot x}}}{32 - \frac{9}{x}}$

Step 10.3

Rewrite the expression.

$6 lim_{x \to - \infty} \frac{- \sqrt{\frac{16 x - 9}{x}}}{32 - \frac{9}{x}}$

Step 11

Evaluate the limit.

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Step 11.1

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

$6 \frac{lim_{x \to - \infty} - \sqrt{\frac{16 x - 9}{x}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 11.2

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

$6 \frac{- lim_{x \to - \infty} \sqrt{\frac{16 x - 9}{x}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 11.3

Move the limit under the radical sign.

$6 \frac{- \sqrt{lim_{x \to - \infty} \frac{16 x - 9}{x}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 12

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

$6 \frac{- \sqrt{lim_{x \to - \infty} \frac{\frac{16 x}{x} + \frac{- 9}{x}}{\frac{x}{x}}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13

Evaluate the limit.

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Step 13.1

Simplify each term.

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Step 13.1.1

Cancel the common factor of $x$ .

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Step 13.1.1.1

Cancel the common factor.

$6 \frac{- \sqrt{lim_{x \to - \infty} \frac{\frac{16 x}{x} + \frac{- 9}{x}}{\frac{x}{x}}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13.1.1.2

Divide $16$ by $1$ .

$6 \frac{- \sqrt{lim_{x \to - \infty} \frac{16 + \frac{- 9}{x}}{\frac{x}{x}}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13.1.2

Move the negative in front of the fraction.

$6 \frac{- \sqrt{lim_{x \to - \infty} \frac{16 - \frac{9}{x}}{\frac{x}{x}}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13.2

Cancel the common factor of $x$ .

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Step 13.2.1

Cancel the common factor.

$6 \frac{- \sqrt{lim_{x \to - \infty} \frac{16 - \frac{9}{x}}{\frac{x}{x}}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13.2.2

Rewrite the expression.

$6 \frac{- \sqrt{lim_{x \to - \infty} \frac{16 - \frac{9}{x}}{1}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13.3

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

$6 \frac{- \sqrt{\frac{lim_{x \to - \infty} 16 - \frac{9}{x}}{lim_{x \to - \infty} 1}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13.4

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

$6 \frac{- \sqrt{\frac{lim_{x \to - \infty} 16 - lim_{x \to - \infty} \frac{9}{x}}{lim_{x \to - \infty} 1}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13.5

Evaluate the limit of $16$ which is constant as $x$ approaches $- \infty$ .

$6 \frac{- \sqrt{\frac{16 - lim_{x \to - \infty} \frac{9}{x}}{lim_{x \to - \infty} 1}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 13.6

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

$6 \frac{- \sqrt{\frac{16 - 9 lim_{x \to - \infty} \frac{1}{x}}{lim_{x \to - \infty} 1}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 14

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

$6 \frac{- \sqrt{\frac{16 - 9 \cdot 0}{lim_{x \to - \infty} 1}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 15

Evaluate the limit.

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Step 15.1

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

$6 \frac{- \sqrt{\frac{16 - 9 \cdot 0}{1}}}{lim_{x \to - \infty} 32 - \frac{9}{x}}$

Step 15.2

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

$6 \frac{- \sqrt{\frac{16 - 9 \cdot 0}{1}}}{lim_{x \to - \infty} 32 - lim_{x \to - \infty} \frac{9}{x}}$

Step 15.3

Evaluate the limit of $32$ which is constant as $x$ approaches $- \infty$ .

$6 \frac{- \sqrt{\frac{16 - 9 \cdot 0}{1}}}{32 - lim_{x \to - \infty} \frac{9}{x}}$

Step 15.4

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

$6 \frac{- \sqrt{\frac{16 - 9 \cdot 0}{1}}}{32 - 9 lim_{x \to - \infty} \frac{1}{x}}$

Step 16

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

$6 \frac{- \sqrt{\frac{16 - 9 \cdot 0}{1}}}{32 - 9 \cdot 0}$

Step 17

Simplify the answer.

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Step 17.1

Divide $16 - 9 \cdot 0$ by $1$ .

$6 \frac{- \sqrt{16 - 9 \cdot 0}}{32 - 9 \cdot 0}$

Step 17.2

Simplify the numerator.

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Step 17.2.1

Multiply $- 9$ by $0$ .

$6 \frac{- \sqrt{16 + 0}}{32 - 9 \cdot 0}$

Step 17.2.2

Add $16$ and $0$ .

$6 \frac{- \sqrt{16}}{32 - 9 \cdot 0}$

Step 17.2.3

Rewrite $16$ as $4^{2}$ .

$6 \frac{- \sqrt{4^{2}}}{32 - 9 \cdot 0}$

Step 17.2.4

Pull terms out from under the radical, assuming positive real numbers.

$6 \frac{- 1 \cdot 4}{32 - 9 \cdot 0}$

Step 17.3

Simplify the denominator.

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Step 17.3.1

Multiply $- 9$ by $0$ .

$6 \frac{- 1 \cdot 4}{32 + 0}$

Step 17.3.2

Add $32$ and $0$ .

$6 \frac{- 1 \cdot 4}{32}$

Step 17.4

Multiply $- 1$ by $4$ .

$6 (\frac{- 4}{32})$

Step 17.5

Cancel the common factor of $2$ .

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Step 17.5.1

Factor $2$ out of $6$ .

$2 (3) \frac{- 4}{32}$

Step 17.5.2

Factor $2$ out of $32$ .

$2 \cdot 3 \frac{- 4}{2 \cdot 16}$

Step 17.5.3

Cancel the common factor.

$2 \cdot 3 \frac{- 4}{2 \cdot 16}$

Step 17.5.4

Rewrite the expression.

$3 (\frac{- 4}{16})$

Step 17.6

Combine $3$ and $\frac{- 4}{16}$ .

$\frac{3 \cdot - 4}{16}$

Step 17.7

Multiply $3$ by $- 4$ .

$\frac{- 12}{16}$

Step 17.8

Cancel the common factor of $- 12$ and $16$ .

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Step 17.8.1

Factor $4$ out of $- 12$ .

$\frac{4 (- 3)}{16}$

Step 17.8.2

Cancel the common factors.

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Step 17.8.2.1

Factor $4$ out of $16$ .

$\frac{4 \cdot - 3}{4 \cdot 4}$

Step 17.8.2.2

Cancel the common factor.

$\frac{4 \cdot - 3}{4 \cdot 4}$

Step 17.8.2.3

Rewrite the expression.

$\frac{- 3}{4}$

Step 17.9

Move the negative in front of the fraction.

$- \frac{3}{4}$