Calculus Examples

$lim_{x \to \infty} \frac{(3 x + 9) (2 x + 7)}{(x + 1) (5 x + 4)}$

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.

$\frac{lim_{x \to \infty} (3 x + 9) (2 x + 7)}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Apply the distributive property.

$\frac{lim_{x \to \infty} 3 x (2 x + 7) + 9 (2 x + 7)}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.2

Apply the distributive property.

$\frac{lim_{x \to \infty} 3 x (2 x) + 3 x \cdot 7 + 9 (2 x + 7)}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.3

Apply the distributive property.

$\frac{lim_{x \to \infty} 3 x (2 x) + 3 x \cdot 7 + 9 (2 x) + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.4

Simplify the expression.

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Step 1.1.2.4.1

Move $x$ .

$\frac{lim_{x \to \infty} 3 \cdot 2 x \cdot x + 3 x \cdot 7 + 9 (2 x) + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.4.2

Move $x$ .

$\frac{lim_{x \to \infty} 3 \cdot 2 x \cdot x + 3 \cdot 7 x + 9 (2 x) + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.4.3

Multiply $3$ by $2$ .

$\frac{lim_{x \to \infty} 6 x \cdot x + 3 \cdot 7 x + 9 \cdot 2 x + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.5

Raise $x$ to the power of $1$ .

$\frac{lim_{x \to \infty} 6 (x^{1} x) + 3 \cdot 7 x + 9 \cdot 2 x + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.6

Raise $x$ to the power of $1$ .

$\frac{lim_{x \to \infty} 6 (x^{1} x^{1}) + 3 \cdot 7 x + 9 \cdot 2 x + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{lim_{x \to \infty} 6 x^{1 + 1} + 3 \cdot 7 x + 9 \cdot 2 x + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.8

Simplify by adding terms.

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Step 1.1.2.8.1

Add $1$ and $1$ .

$\frac{lim_{x \to \infty} 6 x^{2} + 3 \cdot 7 x + 9 \cdot 2 x + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.8.2

Multiply.

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Step 1.1.2.8.2.1

Multiply $3$ by $7$ .

$\frac{lim_{x \to \infty} 6 x^{2} + 21 x + 9 \cdot 2 x + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.8.2.2

Multiply $9$ by $2$ .

$\frac{lim_{x \to \infty} 6 x^{2} + 21 x + 18 x + 9 \cdot 7}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.8.2.3

Multiply $9$ by $7$ .

$\frac{lim_{x \to \infty} 6 x^{2} + 21 x + 18 x + 63}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.8.3

Add $21 x$ and $18 x$ .

$\frac{lim_{x \to \infty} 6 x^{2} + 39 x + 63}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.2.9

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

$\frac{\infty}{lim_{x \to \infty} (x + 1) (5 x + 4)}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Apply the distributive property.

$\frac{\infty}{lim_{x \to \infty} x (5 x + 4) + 1 (5 x + 4)}$

Step 1.1.3.2

Apply the distributive property.

$\frac{\infty}{lim_{x \to \infty} x (5 x) + x \cdot 4 + 1 (5 x + 4)}$

Step 1.1.3.3

Apply the distributive property.

$\frac{\infty}{lim_{x \to \infty} x (5 x) + x \cdot 4 + 1 (5 x) + 1 \cdot 4}$

Step 1.1.3.4

Simplify with commuting.

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Step 1.1.3.4.1

Reorder $x$ and $5$ .

$\frac{\infty}{lim_{x \to \infty} 5 \cdot x \cdot x + x \cdot 4 + 1 (5 x) + 1 \cdot 4}$

Step 1.1.3.4.2

Reorder $x$ and $4$ .

$\frac{\infty}{lim_{x \to \infty} 5 \cdot x \cdot x + 4 \cdot x + 1 (5 x) + 1 \cdot 4}$

Step 1.1.3.5

Raise $x$ to the power of $1$ .

$\frac{\infty}{lim_{x \to \infty} 5 (x^{1} x) + 4 x + 1 \cdot 5 x + 1 \cdot 4}$

Step 1.1.3.6

Raise $x$ to the power of $1$ .

$\frac{\infty}{lim_{x \to \infty} 5 (x^{1} x^{1}) + 4 x + 1 \cdot 5 x + 1 \cdot 4}$

Step 1.1.3.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\infty}{lim_{x \to \infty} 5 x^{1 + 1} + 4 x + 1 \cdot 5 x + 1 \cdot 4}$

Step 1.1.3.8

Simplify by adding terms.

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Step 1.1.3.8.1

Add $1$ and $1$ .

$\frac{\infty}{lim_{x \to \infty} 5 x^{2} + 4 x + 1 \cdot 5 x + 1 \cdot 4}$

Step 1.1.3.8.2

Simplify.

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Step 1.1.3.8.2.1

Multiply $5$ by $1$ .

$\frac{\infty}{lim_{x \to \infty} 5 x^{2} + 4 x + 5 x + 1 \cdot 4}$

Step 1.1.3.8.2.2

Multiply $4$ by $1$ .

$\frac{\infty}{lim_{x \to \infty} 5 x^{2} + 4 x + 5 x + 4}$

Step 1.1.3.8.3

Add $4 x$ and $5 x$ .

$\frac{\infty}{lim_{x \to \infty} 5 x^{2} + 9 x + 4}$

Step 1.1.3.9

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

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

Step 1.1.3.10

Infinity divided by infinity is undefined.

Undefined

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

Step 1.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 1.2

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.

$lim_{x \to \infty} \frac{(3 x + 9) (2 x + 7)}{(x + 1) (5 x + 4)} = lim_{x \to \infty} \frac{\frac{d}{d x} [(3 x + 9) (2 x + 7)]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [(3 x + 9) (2 x + 7)]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 3 x + 9$ and $g (x) = 2 x + 7$ .

$lim_{x \to \infty} \frac{(3 x + 9) \frac{d}{d x} [2 x + 7] + (2 x + 7) \frac{d}{d x} [3 x + 9]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.3

By the Sum Rule, the derivative of $2 x + 7$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [7]$ .

$lim_{x \to \infty} \frac{(3 x + 9) (\frac{d}{d x} [2 x] + \frac{d}{d x} [7]) + (2 x + 7) \frac{d}{d x} [3 x + 9]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.4

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

$lim_{x \to \infty} \frac{(3 x + 9) (2 \frac{d}{d x} [x] + \frac{d}{d x} [7]) + (2 x + 7) \frac{d}{d x} [3 x + 9]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.5

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

$lim_{x \to \infty} \frac{(3 x + 9) (2 \cdot 1 + \frac{d}{d x} [7]) + (2 x + 7) \frac{d}{d x} [3 x + 9]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.6

Multiply $2$ by $1$ .

$lim_{x \to \infty} \frac{(3 x + 9) (2 + \frac{d}{d x} [7]) + (2 x + 7) \frac{d}{d x} [3 x + 9]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.7

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{(3 x + 9) (2 + 0) + (2 x + 7) \frac{d}{d x} [3 x + 9]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.8

Add $2$ and $0$ .

$lim_{x \to \infty} \frac{(3 x + 9) \cdot 2 + (2 x + 7) \frac{d}{d x} [3 x + 9]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.9

Move $2$ to the left of $3 x + 9$ .

$lim_{x \to \infty} \frac{2 \cdot (3 x + 9) + (2 x + 7) \frac{d}{d x} [3 x + 9]}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.10

By the Sum Rule, the derivative of $3 x + 9$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [9]$ .

$lim_{x \to \infty} \frac{2 (3 x + 9) + (2 x + 7) (\frac{d}{d x} [3 x] + \frac{d}{d x} [9])}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.11

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

$lim_{x \to \infty} \frac{2 (3 x + 9) + (2 x + 7) (3 \frac{d}{d x} [x] + \frac{d}{d x} [9])}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.12

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

$lim_{x \to \infty} \frac{2 (3 x + 9) + (2 x + 7) (3 \cdot 1 + \frac{d}{d x} [9])}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.13

Multiply $3$ by $1$ .

$lim_{x \to \infty} \frac{2 (3 x + 9) + (2 x + 7) (3 + \frac{d}{d x} [9])}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.14

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{2 (3 x + 9) + (2 x + 7) (3 + 0)}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.15

Add $3$ and $0$ .

$lim_{x \to \infty} \frac{2 (3 x + 9) + (2 x + 7) \cdot 3}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.16

Move $3$ to the left of $2 x + 7$ .

$lim_{x \to \infty} \frac{2 (3 x + 9) + 3 \cdot (2 x + 7)}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.17

Simplify.

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Step 1.3.17.1

Apply the distributive property.

$lim_{x \to \infty} \frac{2 (3 x) + 2 \cdot 9 + 3 (2 x + 7)}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.17.2

Apply the distributive property.

$lim_{x \to \infty} \frac{2 (3 x) + 2 \cdot 9 + 3 (2 x) + 3 \cdot 7}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.17.3

Combine terms.

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Step 1.3.17.3.1

Multiply $3$ by $2$ .

$lim_{x \to \infty} \frac{6 x + 2 \cdot 9 + 3 (2 x) + 3 \cdot 7}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.17.3.2

Multiply $2$ by $9$ .

$lim_{x \to \infty} \frac{6 x + 18 + 3 (2 x) + 3 \cdot 7}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.17.3.3

Multiply $2$ by $3$ .

$lim_{x \to \infty} \frac{6 x + 18 + 6 x + 3 \cdot 7}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.17.3.4

Multiply $3$ by $7$ .

$lim_{x \to \infty} \frac{6 x + 18 + 6 x + 21}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.17.3.5

Add $6 x$ and $6 x$ .

$lim_{x \to \infty} \frac{12 x + 18 + 21}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.17.3.6

Add $18$ and $21$ .

$lim_{x \to \infty} \frac{12 x + 39}{\frac{d}{d x} [(x + 1) (5 x + 4)]}$

Step 1.3.18

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 1$ and $g (x) = 5 x + 4$ .

$lim_{x \to \infty} \frac{12 x + 39}{(x + 1) \frac{d}{d x} [5 x + 4] + (5 x + 4) \frac{d}{d x} [x + 1]}$

Step 1.3.19

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

$lim_{x \to \infty} \frac{12 x + 39}{(x + 1) (\frac{d}{d x} [5 x] + \frac{d}{d x} [4]) + (5 x + 4) \frac{d}{d x} [x + 1]}$

Step 1.3.20

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

$lim_{x \to \infty} \frac{12 x + 39}{(x + 1) (5 \frac{d}{d x} [x] + \frac{d}{d x} [4]) + (5 x + 4) \frac{d}{d x} [x + 1]}$

Step 1.3.21

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

$lim_{x \to \infty} \frac{12 x + 39}{(x + 1) (5 \cdot 1 + \frac{d}{d x} [4]) + (5 x + 4) \frac{d}{d x} [x + 1]}$

Step 1.3.22

Multiply $5$ by $1$ .

$lim_{x \to \infty} \frac{12 x + 39}{(x + 1) (5 + \frac{d}{d x} [4]) + (5 x + 4) \frac{d}{d x} [x + 1]}$

Step 1.3.23

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{12 x + 39}{(x + 1) (5 + 0) + (5 x + 4) \frac{d}{d x} [x + 1]}$

Step 1.3.24

Add $5$ and $0$ .

$lim_{x \to \infty} \frac{12 x + 39}{(x + 1) \cdot 5 + (5 x + 4) \frac{d}{d x} [x + 1]}$

Step 1.3.25

Move $5$ to the left of $x + 1$ .

$lim_{x \to \infty} \frac{12 x + 39}{5 \cdot (x + 1) + (5 x + 4) \frac{d}{d x} [x + 1]}$

Step 1.3.26

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$lim_{x \to \infty} \frac{12 x + 39}{5 (x + 1) + (5 x + 4) (\frac{d}{d x} [x] + \frac{d}{d x} [1])}$

Step 1.3.27

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

$lim_{x \to \infty} \frac{12 x + 39}{5 (x + 1) + (5 x + 4) (1 + \frac{d}{d x} [1])}$

Step 1.3.28

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{12 x + 39}{5 (x + 1) + (5 x + 4) (1 + 0)}$

Step 1.3.29

Add $1$ and $0$ .

$lim_{x \to \infty} \frac{12 x + 39}{5 (x + 1) + (5 x + 4) \cdot 1}$

Step 1.3.30

Multiply $5 x + 4$ by $1$ .

$lim_{x \to \infty} \frac{12 x + 39}{5 (x + 1) + 5 x + 4}$

Step 1.3.31

Simplify.

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Step 1.3.31.1

Apply the distributive property.

$lim_{x \to \infty} \frac{12 x + 39}{5 x + 5 \cdot 1 + 5 x + 4}$

Step 1.3.31.2

Combine terms.

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Step 1.3.31.2.1

Multiply $5$ by $1$ .

$lim_{x \to \infty} \frac{12 x + 39}{5 x + 5 + 5 x + 4}$

Step 1.3.31.2.2

Add $5 x$ and $5 x$ .

$lim_{x \to \infty} \frac{12 x + 39}{10 x + 5 + 4}$

Step 1.3.31.2.3

Add $5$ and $4$ .

$lim_{x \to \infty} \frac{12 x + 39}{10 x + 9}$

Step 2

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

$lim_{x \to \infty} \frac{\frac{12 x}{x} + \frac{39}{x}}{\frac{10 x}{x} + \frac{9}{x}}$

Step 3

Evaluate the limit.

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Step 3.1

Cancel the common factor of $x$ .

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Step 3.1.1

Cancel the common factor.

$lim_{x \to \infty} \frac{\frac{12 x}{x} + \frac{39}{x}}{\frac{10 x}{x} + \frac{9}{x}}$

Step 3.1.2

Divide $12$ by $1$ .

$lim_{x \to \infty} \frac{12 + \frac{39}{x}}{\frac{10 x}{x} + \frac{9}{x}}$

Step 3.2

Cancel the common factor of $x$ .

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Step 3.2.1

Cancel the common factor.

$lim_{x \to \infty} \frac{12 + \frac{39}{x}}{\frac{10 x}{x} + \frac{9}{x}}$

Step 3.2.2

Divide $10$ by $1$ .

$lim_{x \to \infty} \frac{12 + \frac{39}{x}}{10 + \frac{9}{x}}$

Step 3.3

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

$\frac{lim_{x \to \infty} 12 + \frac{39}{x}}{lim_{x \to \infty} 10 + \frac{9}{x}}$

Step 3.4

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

$\frac{lim_{x \to \infty} 12 + lim_{x \to \infty} \frac{39}{x}}{lim_{x \to \infty} 10 + \frac{9}{x}}$

Step 3.5

Evaluate the limit of $12$ which is constant as $x$ approaches $\infty$ .

$\frac{12 + lim_{x \to \infty} \frac{39}{x}}{lim_{x \to \infty} 10 + \frac{9}{x}}$

Step 3.6

Move the term $39$ outside of the limit because it is constant with respect to $x$ .

$\frac{12 + 39 lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} 10 + \frac{9}{x}}$

Step 4

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

$\frac{12 + 39 \cdot 0}{lim_{x \to \infty} 10 + \frac{9}{x}}$

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 $\infty$ .

$\frac{12 + 39 \cdot 0}{lim_{x \to \infty} 10 + lim_{x \to \infty} \frac{9}{x}}$

Step 5.2

Evaluate the limit of $10$ which is constant as $x$ approaches $\infty$ .

$\frac{12 + 39 \cdot 0}{10 + lim_{x \to \infty} \frac{9}{x}}$

Step 5.3

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

$\frac{12 + 39 \cdot 0}{10 + 9 lim_{x \to \infty} \frac{1}{x}}$

Step 6

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

$\frac{12 + 39 \cdot 0}{10 + 9 \cdot 0}$

Step 7

Simplify the answer.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Multiply $39$ by $0$ .

$\frac{12 + 0}{10 + 9 \cdot 0}$

Step 7.1.2

Add $12$ and $0$ .

$\frac{12}{10 + 9 \cdot 0}$

Step 7.2

Simplify the denominator.

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Step 7.2.1

Multiply $9$ by $0$ .

$\frac{12}{10 + 0}$

Step 7.2.2

Add $10$ and $0$ .

$\frac{12}{10}$

Step 7.3

Cancel the common factor of $12$ and $10$ .

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Step 7.3.1

Factor $2$ out of $12$ .

$\frac{2 (6)}{10}$

Step 7.3.2

Cancel the common factors.

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Step 7.3.2.1

Factor $2$ out of $10$ .

$\frac{2 \cdot 6}{2 \cdot 5}$

Step 7.3.2.2

Cancel the common factor.

$\frac{2 \cdot 6}{2 \cdot 5}$

Step 7.3.2.3

Rewrite the expression.

$\frac{6}{5}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{6}{5}$

Decimal Form:

$1.2$