Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Apply the distributive property.

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

Step 1.1.2.2

Apply the distributive property.

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

Step 1.1.2.3

Apply the distributive property.

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

Step 1.1.2.4

Simplify the expression.

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Step 1.1.2.4.1

Move $x$ .

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

Step 1.1.2.4.2

Move $x$ .

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

Step 1.1.2.4.3

Multiply $4$ by $2$ .

$\frac{lim_{x \to \infty} 8 + 4 \cdot - 1 x + 5 \cdot 2 x + 5 \cdot - 1 x \cdot x}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.4.4

Multiply $4$ by $- 1$ .

$\frac{lim_{x \to \infty} 8 - 4 x + 5 \cdot 2 x + 5 \cdot - 1 x \cdot x}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.4.5

Multiply $5$ by $2$ .

$\frac{lim_{x \to \infty} 8 - 4 x + 10 x + 5 \cdot - 1 x \cdot x}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.4.6

Multiply $5$ by $- 1$ .

$\frac{lim_{x \to \infty} 8 - 4 x + 10 x - 5 x \cdot x}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.5

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

$\frac{lim_{x \to \infty} 8 - 4 x + 10 x - 5 (x^{1} x)}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.6

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

$\frac{lim_{x \to \infty} 8 - 4 x + 10 x - 5 (x^{1} x^{1})}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.7

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

$\frac{lim_{x \to \infty} 8 - 4 x + 10 x - 5 x^{1 + 1}}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.8

Simplify by adding terms.

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Step 1.1.2.8.1

Add $1$ and $1$ .

$\frac{lim_{x \to \infty} 8 - 4 x + 10 x - 5 x^{2}}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.8.2

Add $- 4 x$ and $10 x$ .

$\frac{lim_{x \to \infty} 8 + 6 x - 5 x^{2}}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.8.3

Simplify the expression.

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Step 1.1.2.8.3.1

Reorder $6 x$ and $- 5 x^{2}$ .

$\frac{lim_{x \to \infty} 8 - 5 x^{2} + 6 x}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.8.3.2

Move $8$ .

$\frac{lim_{x \to \infty} - 5 x^{2} + 6 x + 8}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.2.9

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

$\frac{- \infty}{lim_{x \to \infty} (2 + x) (1 - x)}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Apply the distributive property.

$\frac{- \infty}{lim_{x \to \infty} 2 (1 - x) + x (1 - x)}$

Step 1.1.3.2

Apply the distributive property.

$\frac{- \infty}{lim_{x \to \infty} 2 \cdot 1 + 2 (- x) + x (1 - x)}$

Step 1.1.3.3

Apply the distributive property.

$\frac{- \infty}{lim_{x \to \infty} 2 \cdot 1 + 2 (- x) + x \cdot 1 + x (- x)}$

Step 1.1.3.4

Simplify the expression.

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Step 1.1.3.4.1

Reorder $x$ and $1$ .

$\frac{- \infty}{lim_{x \to \infty} 2 \cdot 1 + 2 \cdot - 1 x + 1 \cdot x + x (- x)}$

Step 1.1.3.4.2

Reorder $x$ and $- 1$ .

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

Step 1.1.3.4.3

Multiply $2$ by $1$ .

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

Step 1.1.3.4.4

Multiply $2$ by $- 1$ .

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

Step 1.1.3.4.5

Multiply $x$ by $1$ .

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

Step 1.1.3.5

Factor out negative.

$\frac{- \infty}{lim_{x \to \infty} 2 - 2 x + x - (x \cdot x)}$

Step 1.1.3.6

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

$\frac{- \infty}{lim_{x \to \infty} 2 - 2 x + x - (x^{1} x)}$

Step 1.1.3.7

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

$\frac{- \infty}{lim_{x \to \infty} 2 - 2 x + x - (x^{1} x^{1})}$

Step 1.1.3.8

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

$\frac{- \infty}{lim_{x \to \infty} 2 - 2 x + x - x^{1 + 1}}$

Step 1.1.3.9

Simplify by adding terms.

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Step 1.1.3.9.1

Add $1$ and $1$ .

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

Step 1.1.3.9.2

Add $- 2 x$ and $x$ .

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

Step 1.1.3.9.3

Simplify the expression.

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Step 1.1.3.9.3.1

Reorder $- x$ and $- x^{2}$ .

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

Step 1.1.3.9.3.2

Move $2$ .

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

Step 1.1.3.10

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

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

Step 1.1.3.11

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

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 $f (x) = 4 + 5 x$ and $g (x) = 2 - x$ .

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

Step 1.3.3

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

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

Step 1.3.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

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

Step 1.3.5

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 1.3.6

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 1.3.7

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

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

Step 1.3.8

Multiply $- 1$ by $1$ .

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

Step 1.3.9

Move $- 1$ to the left of $4 + 5 x$ .

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

Step 1.3.10

Rewrite $- 1 (4 + 5 x)$ as $- (4 + 5 x)$ .

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

Step 1.3.11

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

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

Step 1.3.12

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

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

Step 1.3.13

Add $0$ and $\frac{d}{d x} [5 x]$ .

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

Step 1.3.14

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

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

Step 1.3.15

Move $5$ to the left of $2 - x$ .

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

Step 1.3.16

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

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

Step 1.3.17

Multiply $5$ by $1$ .

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

Step 1.3.18

Simplify.

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Step 1.3.18.1

Apply the distributive property.

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

Step 1.3.18.2

Apply the distributive property.

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

Step 1.3.18.3

Combine terms.

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Step 1.3.18.3.1

Multiply $- 1$ by $4$ .

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

Step 1.3.18.3.2

Multiply $5$ by $- 1$ .

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

Step 1.3.18.3.3

Multiply $5$ by $2$ .

$lim_{x \to \infty} \frac{- 4 - 5 x + 10 + 5 (- x)}{\frac{d}{d x} [(2 + x) (1 - x)]}$

Step 1.3.18.3.4

Multiply $- 1$ by $5$ .

$lim_{x \to \infty} \frac{- 4 - 5 x + 10 - 5 x}{\frac{d}{d x} [(2 + x) (1 - x)]}$

Step 1.3.18.3.5

Add $- 4$ and $10$ .

$lim_{x \to \infty} \frac{- 5 x + 6 - 5 x}{\frac{d}{d x} [(2 + x) (1 - x)]}$

Step 1.3.18.3.6

Subtract $5 x$ from $- 5 x$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{\frac{d}{d x} [(2 + x) (1 - x)]}$

Step 1.3.19

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 $f (x) = 2 + x$ and $g (x) = 1 - x$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{(2 + x) \frac{d}{d x} [1 - x] + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.20

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

$lim_{x \to \infty} \frac{- 10 x + 6}{(2 + x) (\frac{d}{d x} [1] + \frac{d}{d x} [- x]) + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.21

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

$lim_{x \to \infty} \frac{- 10 x + 6}{(2 + x) (0 + \frac{d}{d x} [- x]) + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.22

Add $0$ and $\frac{d}{d x} [- x]$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{(2 + x) \frac{d}{d x} [- x] + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.23

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{(2 + x) (- \frac{d}{d x} [x]) + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.24

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

$lim_{x \to \infty} \frac{- 10 x + 6}{(2 + x) (- 1 \cdot 1) + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.25

Multiply $- 1$ by $1$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{(2 + x) \cdot - 1 + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.26

Move $- 1$ to the left of $2 + x$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{- 1 \cdot (2 + x) + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.27

Rewrite $- 1 (2 + x)$ as $- (2 + x)$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{- (2 + x) + (1 - x) \frac{d}{d x} [2 + x]}$

Step 1.3.28

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

$lim_{x \to \infty} \frac{- 10 x + 6}{- (2 + x) + (1 - x) (\frac{d}{d x} [2] + \frac{d}{d x} [x])}$

Step 1.3.29

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{- (2 + x) + (1 - x) (0 + \frac{d}{d x} [x])}$

Step 1.3.30

Add $0$ and $\frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{- (2 + x) + (1 - x) \frac{d}{d x} [x]}$

Step 1.3.31

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

$lim_{x \to \infty} \frac{- 10 x + 6}{- (2 + x) + (1 - x) \cdot 1}$

Step 1.3.32

Multiply $1 - x$ by $1$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{- (2 + x) + 1 - x}$

Step 1.3.33

Simplify.

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Step 1.3.33.1

Apply the distributive property.

$lim_{x \to \infty} \frac{- 10 x + 6}{- 1 \cdot 2 - x + 1 - x}$

Step 1.3.33.2

Combine terms.

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Step 1.3.33.2.1

Multiply $- 1$ by $2$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{- 2 - x + 1 - x}$

Step 1.3.33.2.2

Add $- 2$ and $1$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{- x - 1 - x}$

Step 1.3.33.2.3

Subtract $x$ from $- x$ .

$lim_{x \to \infty} \frac{- 10 x + 6}{- 2 x - 1}$

Step 2

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

$lim_{x \to \infty} \frac{\frac{- 10 x}{x} + \frac{6}{x}}{\frac{- 2 x}{x} - \frac{1}{x}}$

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.

$lim_{x \to \infty} \frac{\frac{- 10 x}{x} + \frac{6}{x}}{\frac{- 2 x}{x} - \frac{1}{x}}$

Step 3.1.2

Divide $- 10$ by $1$ .

$lim_{x \to \infty} \frac{- 10 + \frac{6}{x}}{\frac{- 2 x}{x} - \frac{1}{x}}$

Step 3.2

Cancel the common factor of $x$ .

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Step 3.2.1

Cancel the common factor.

$lim_{x \to \infty} \frac{- 10 + \frac{6}{x}}{\frac{- 2 x}{x} - \frac{1}{x}}$

Step 3.2.2

Divide $- 2$ by $1$ .

$lim_{x \to \infty} \frac{- 10 + \frac{6}{x}}{- 2 - \frac{1}{x}}$

Step 3.3

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

$\frac{lim_{x \to \infty} - 10 + \frac{6}{x}}{lim_{x \to \infty} - 2 - \frac{1}{x}}$

Step 3.4

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

$\frac{- lim_{x \to \infty} 10 + lim_{x \to \infty} \frac{6}{x}}{lim_{x \to \infty} - 2 - \frac{1}{x}}$

Step 3.5

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

$\frac{- 1 \cdot 10 + lim_{x \to \infty} \frac{6}{x}}{lim_{x \to \infty} - 2 - \frac{1}{x}}$

Step 3.6

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

$\frac{- 10 + 6 lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} - 2 - \frac{1}{x}}$

Step 4

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

$\frac{- 10 + 6 \cdot 0}{lim_{x \to \infty} - 2 - \frac{1}{x}}$

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 $\infty$ .

$\frac{- 10 + 6 \cdot 0}{- lim_{x \to \infty} 2 - lim_{x \to \infty} \frac{1}{x}}$

Step 5.2

Evaluate the limit of $2$ which is constant as $x$ approaches $\infty$ .

$\frac{- 10 + 6 \cdot 0}{- 1 \cdot 2 - lim_{x \to \infty} \frac{1}{x}}$

Step 6

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

$\frac{- 10 + 6 \cdot 0}{- 2 - 0}$

Step 7

Simplify the answer.

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Step 7.1

Cancel the common factor of $- 10 + 6 \cdot 0$ and $- 2 - 0$ .

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Step 7.1.1

Rewrite $- 2$ as $- 1 (2)$ .

$\frac{- 10 + 6 \cdot 0}{- 1 (2) - 0}$

Step 7.1.2

Factor $- 1$ out of $- 0$ .

$\frac{- 10 + 6 \cdot 0}{- 1 (2) - (0)}$

Step 7.1.3

Factor $- 1$ out of $- 1 (2) - (0)$ .

$\frac{- 10 + 6 \cdot 0}{- 1 (2 + 0)}$

Step 7.1.4

Factor $2$ out of $- 10$ .

$\frac{2 (- 5) + 6 \cdot 0}{- 1 (2 + 0)}$

Step 7.1.5

Factor $2$ out of $6 \cdot 0$ .

$\frac{2 (- 5) + 2 (3 \cdot 0)}{- 1 (2 + 0)}$

Step 7.1.6

Factor $2$ out of $2 (- 5) + 2 (3 \cdot 0)$ .

$\frac{2 (- 5 + 3 \cdot 0)}{- 1 (2 + 0)}$

Step 7.1.7

Cancel the common factors.

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Step 7.1.7.1

Factor $2$ out of $- 1 (2 + 0)$ .

$\frac{2 (- 5 + 3 \cdot 0)}{2 (- 1 (1 + 0))}$

Step 7.1.7.2

Cancel the common factor.

$\frac{2 (- 5 + 3 \cdot 0)}{2 (- 1 (1 + 0))}$

Step 7.1.7.3

Rewrite the expression.

$\frac{- 5 + 3 \cdot 0}{- 1 (1 + 0)}$

Step 7.2

Simplify the numerator.

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Step 7.2.1

Multiply $3$ by $0$ .

$\frac{- 5 + 0}{- 1 (1 + 0)}$

Step 7.2.2

Add $- 5$ and $0$ .

$\frac{- 5}{- 1 (1 + 0)}$

Step 7.3

Add $1$ and $0$ .

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

Step 7.4

Multiply $- 1$ by $1$ .

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

Step 7.5

Divide $- 5$ by $- 1$ .

$5$