Calculus Examples

$lim_{x \to \infty} 3 - x + \frac{x^{2} - 2 x}{x + 5}$

Step 1

Combine terms.

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Step 1.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$lim_{x \to \infty} - x + \frac{3 (x + 5)}{x + 5} + \frac{x^{2} - 2 x}{x + 5}$

Step 1.2

Combine the numerators over the common denominator.

$lim_{x \to \infty} - x + \frac{3 (x + 5) + x^{2} - 2 x}{x + 5}$

Step 1.3

To write $- x$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$lim_{x \to \infty} - x \cdot \frac{x + 5}{x + 5} + \frac{3 (x + 5) + x^{2} - 2 x}{x + 5}$

Step 1.4

Combine $- x$ and $\frac{x + 5}{x + 5}$ .

$lim_{x \to \infty} \frac{- x (x + 5)}{x + 5} + \frac{3 (x + 5) + x^{2} - 2 x}{x + 5}$

Step 1.5

Combine the numerators over the common denominator.

$lim_{x \to \infty} \frac{- x (x + 5) + 3 (x + 5) + x^{2} - 2 x}{x + 5}$

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.

$\frac{lim_{x \to \infty} - x (x + 5) + 3 (x + 5) + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Apply the distributive property.

$\frac{lim_{x \to \infty} - x \cdot x - x \cdot 5 + 3 (x + 5) + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.2

Apply the distributive property.

$\frac{lim_{x \to \infty} - x \cdot x - x \cdot 5 + 3 x + 3 \cdot 5 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.3

Move $x$ .

$\frac{lim_{x \to \infty} - x \cdot x - 1 \cdot 5 x + 3 x + 3 \cdot 5 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.4

Factor out negative.

$\frac{lim_{x \to \infty} - (x \cdot x) - 1 \cdot 5 x + 3 x + 3 \cdot 5 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.5

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

$\frac{lim_{x \to \infty} - (x^{1} x) - 1 \cdot 5 x + 3 x + 3 \cdot 5 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.6

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

$\frac{lim_{x \to \infty} - (x^{1} x^{1}) - 1 \cdot 5 x + 3 x + 3 \cdot 5 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.7

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

$\frac{lim_{x \to \infty} - x^{1 + 1} - 1 \cdot 5 x + 3 x + 3 \cdot 5 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8

Simplify by adding terms.

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Step 2.1.2.8.1

Add $1$ and $1$ .

$\frac{lim_{x \to \infty} - x^{2} - 1 \cdot 5 x + 3 x + 3 \cdot 5 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.2

Multiply.

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Step 2.1.2.8.2.1

Multiply $- 1$ by $5$ .

$\frac{lim_{x \to \infty} - x^{2} - 5 x + 3 x + 3 \cdot 5 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.2.2

Multiply $3$ by $5$ .

$\frac{lim_{x \to \infty} - x^{2} - 5 x + 3 x + 15 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.3

Add $- 5 x$ and $3 x$ .

$\frac{lim_{x \to \infty} - x^{2} - 2 x + 15 + x^{2} - 2 x}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.4

Simplify the expression.

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Step 2.1.2.8.4.1

Move $15$ .

$\frac{lim_{x \to \infty} - x^{2} - 2 x + x^{2} - 2 x + 15}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.4.2

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

$\frac{lim_{x \to \infty} - x^{2} + x^{2} - 2 x - 2 x + 15}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.4.3

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

$\frac{lim_{x \to \infty} x^{2} - x^{2} - 2 x - 2 x + 15}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.5

Subtract $x^{2}$ from $x^{2}$ .

$\frac{lim_{x \to \infty} 0 - 2 x - 2 x + 15}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.6

Subtract $2 x$ from $0$ .

$\frac{lim_{x \to \infty} - 2 x - 2 x + 15}{lim_{x \to \infty} x + 5}$

Step 2.1.2.8.7

Subtract $2 x$ from $- 2 x$ .

$\frac{lim_{x \to \infty} - 4 x + 15}{lim_{x \to \infty} x + 5}$

Step 2.1.2.9

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

$\frac{- \infty}{lim_{x \to \infty} x + 5}$

Step 2.1.3

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

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

Step 2.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 2.2

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

Step 2.3.2

By the Sum Rule, the derivative of $- x (x + 5) + 3 (x + 5) + x^{2} - 2 x$ with respect to $x$ is $\frac{d}{d x} [- x (x + 5)] + \frac{d}{d x} [3 (x + 5)] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x]$ .

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

Step 2.3.3

Evaluate $\frac{d}{d x} [- x (x + 5)]$ .

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Step 2.3.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x (x + 5)$ with respect to $x$ is $- \frac{d}{d x} [x (x + 5)]$ .

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

Step 2.3.3.2

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 $f (x) = x$ and $g (x) = x + 5$ .

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

Step 2.3.3.3

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

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

Step 2.3.3.4

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

$lim_{x \to \infty} \frac{- (x (1 + \frac{d}{d x} [5]) + (x + 5) \frac{d}{d x} [x]) + \frac{d}{d x} [3 (x + 5)] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x]}{\frac{d}{d x} [x + 5]}$

Step 2.3.3.5

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

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

Step 2.3.3.6

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

$lim_{x \to \infty} \frac{- (x (1 + 0) + (x + 5) \cdot 1) + \frac{d}{d x} [3 (x + 5)] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x]}{\frac{d}{d x} [x + 5]}$

Step 2.3.3.7

Add $1$ and $0$ .

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

Step 2.3.3.8

Multiply $x$ by $1$ .

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

Step 2.3.3.9

Multiply $x + 5$ by $1$ .

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

Step 2.3.3.10

Add $x$ and $x$ .

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

Step 2.3.4

Evaluate $\frac{d}{d x} [3 (x + 5)]$ .

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Step 2.3.4.1

Since $3$ is constant with respect to $x$ , the derivative of $3 (x + 5)$ with respect to $x$ is $3 \frac{d}{d x} [x + 5]$ .

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

Step 2.3.4.2

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

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

Step 2.3.4.3

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

$lim_{x \to \infty} \frac{- (2 x + 5) + 3 (1 + \frac{d}{d x} [5]) + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x]}{\frac{d}{d x} [x + 5]}$

Step 2.3.4.4

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

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

Step 2.3.4.5

Add $1$ and $0$ .

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

Step 2.3.4.6

Multiply $3$ by $1$ .

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

Step 2.3.5

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

$lim_{x \to \infty} \frac{- (2 x + 5) + 3 + 2 x + \frac{d}{d x} [- 2 x]}{\frac{d}{d x} [x + 5]}$

Step 2.3.6

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

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Step 2.3.6.1

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

$lim_{x \to \infty} \frac{- (2 x + 5) + 3 + 2 x - 2 \frac{d}{d x} [x]}{\frac{d}{d x} [x + 5]}$

Step 2.3.6.2

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

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

Step 2.3.6.3

Multiply $- 2$ by $1$ .

$lim_{x \to \infty} \frac{- (2 x + 5) + 3 + 2 x - 2}{\frac{d}{d x} [x + 5]}$

Step 2.3.7

Simplify.

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Step 2.3.7.1

Apply the distributive property.

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

Step 2.3.7.2

Combine terms.

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Step 2.3.7.2.1

Multiply $2$ by $- 1$ .

$lim_{x \to \infty} \frac{- 2 x - 1 \cdot 5 + 3 + 2 x - 2}{\frac{d}{d x} [x + 5]}$

Step 2.3.7.2.2

Multiply $- 1$ by $5$ .

$lim_{x \to \infty} \frac{- 2 x - 5 + 3 + 2 x - 2}{\frac{d}{d x} [x + 5]}$

Step 2.3.7.2.3

Add $- 5$ and $3$ .

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

Step 2.3.7.2.4

Add $- 2 x$ and $2 x$ .

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

Step 2.3.7.2.5

Subtract $2$ from $0$ .

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

Step 2.3.7.2.6

Subtract $2$ from $- 2$ .

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

Step 2.3.8

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

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

Step 2.3.9

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

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

Step 2.3.10

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{- 4}{1 + 0}$

Step 2.3.11

Add $1$ and $0$ .

$lim_{x \to \infty} \frac{- 4}{1}$

Step 2.4

Divide $- 4$ by $1$ .

$lim_{x \to \infty} - 4$

Step 3

Evaluate the limit of $- 4$ which is constant as $x$ approaches $\infty$ .

$- 4$