Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Apply basic rules of exponents.

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Step 1.1.2.1.1

Rewrite ${(n - 1)}^{2}$ as $(n - 1) (n - 1)$ .

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

Step 1.1.2.1.2

Rewrite ${(n + 2)}^{2}$ as $(n + 2) (n + 2)$ .

$\frac{lim_{n \to \infty} (n - 1) (n - 1) - ((n + 2) (n + 2))}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.2

Apply the distributive property.

$\frac{lim_{n \to \infty} n (n - 1) - 1 (n - 1) - ((n + 2) (n + 2))}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.3

Apply the distributive property.

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

Step 1.1.2.4

Apply the distributive property.

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

Step 1.1.2.5

Apply the distributive property.

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

Step 1.1.2.6

Apply the distributive property.

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

Step 1.1.2.7

Apply the distributive property.

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

Step 1.1.2.8

Apply the distributive property.

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

Step 1.1.2.9

Apply the distributive property.

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

Step 1.1.2.10

Apply the distributive property.

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

Step 1.1.2.11

Simplify with commuting.

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Step 1.1.2.11.1

Reorder $n$ and $- 1$ .

$\frac{lim_{n \to \infty} n \cdot n - 1 \cdot n - 1 n - 1 \cdot - 1 - (n \cdot n) - (n \cdot 2) - (2 n) - (2 \cdot 2)}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.11.2

Reorder $n$ and $2$ .

$\frac{lim_{n \to \infty} n \cdot n - 1 \cdot n - 1 n - 1 \cdot - 1 - n \cdot n - (2 \cdot n) - (2 n) - (2 \cdot 2)}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.12

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

$\frac{lim_{n \to \infty} n^{1} n - 1 n - 1 n - 1 \cdot - 1 - n \cdot n - 1 \cdot 2 n - 1 \cdot 2 n - 1 \cdot 2 \cdot 2}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.13

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

$\frac{lim_{n \to \infty} n^{1} n^{1} - 1 n - 1 n - 1 \cdot - 1 - n \cdot n - 1 \cdot 2 n - 1 \cdot 2 n - 1 \cdot 2 \cdot 2}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.14

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

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

Step 1.1.2.15

Simplify the expression.

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Step 1.1.2.15.1

Add $1$ and $1$ .

$\frac{lim_{n \to \infty} n^{2} - 1 n - 1 n - 1 \cdot - 1 - n \cdot n - 1 \cdot 2 n - 1 \cdot 2 n - 1 \cdot 2 \cdot 2}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.15.2

Multiply $- 1$ by $- 1$ .

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

Step 1.1.2.16

Subtract $1 n$ from $- 1 n$ .

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

Step 1.1.2.17

Factor out negative.

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

Step 1.1.2.18

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

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

Step 1.1.2.19

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

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

Step 1.1.2.20

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

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

Step 1.1.2.21

Simplify by adding terms.

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Step 1.1.2.21.1

Add $1$ and $1$ .

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

Step 1.1.2.21.2

Multiply.

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Step 1.1.2.21.2.1

Multiply $- 1$ by $2$ .

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

Step 1.1.2.21.2.2

Multiply $- 1$ by $2$ .

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

Step 1.1.2.21.2.3

Multiply $- 1$ by $2$ .

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

Step 1.1.2.21.2.4

Multiply $- 2$ by $2$ .

$\frac{lim_{n \to \infty} n^{2} - 2 n + 1 - n^{2} - 2 n - 2 n - 4}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.21.3

Subtract $2 n$ from $- 2 n$ .

$\frac{lim_{n \to \infty} n^{2} - 2 n + 1 - n^{2} - 4 n - 4}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.21.4

Simplify the expression.

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Step 1.1.2.21.4.1

Move $1$ .

$\frac{lim_{n \to \infty} n^{2} - 2 n - n^{2} - 4 n + 1 - 4}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.21.4.2

Move $- 2 n$ .

$\frac{lim_{n \to \infty} n^{2} - n^{2} - 2 n - 4 n + 1 - 4}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.21.5

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

$\frac{lim_{n \to \infty} 0 - 2 n - 4 n + 1 - 4}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.21.6

Subtract $2 n$ from $0$ .

$\frac{lim_{n \to \infty} - 2 n - 4 n + 1 - 4}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.21.7

Subtract $4 n$ from $- 2 n$ .

$\frac{lim_{n \to \infty} - 6 n + 1 - 4}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.21.8

Subtract $4$ from $1$ .

$\frac{lim_{n \to \infty} - 6 n - 3}{lim_{n \to \infty} 3 - n}$

Step 1.1.2.22

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

$\frac{- \infty}{lim_{n \to \infty} 3 - n}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Reorder $3$ and $- n$ .

$\frac{- \infty}{lim_{n \to \infty} - n + 3}$

Step 1.1.3.2

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

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

Step 1.1.3.3

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

Rewrite ${(n - 1)}^{2}$ as $(n - 1) (n - 1)$ .

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

Step 1.3.3

Expand $(n - 1) (n - 1)$ using the FOIL Method.

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Step 1.3.3.1

Apply the distributive property.

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

Step 1.3.3.2

Apply the distributive property.

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

Step 1.3.3.3

Apply the distributive property.

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

Step 1.3.4

Simplify and combine like terms.

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Step 1.3.4.1

Simplify each term.

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Step 1.3.4.1.1

Multiply $n$ by $n$ .

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

Step 1.3.4.1.2

Move $- 1$ to the left of $n$ .

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

Step 1.3.4.1.3

Rewrite $- 1 n$ as $- n$ .

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

Step 1.3.4.1.4

Rewrite $- 1 n$ as $- n$ .

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

Step 1.3.4.1.5

Multiply $- 1$ by $- 1$ .

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

Step 1.3.4.2

Subtract $n$ from $- n$ .

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

Step 1.3.5

Rewrite ${(n + 2)}^{2}$ as $(n + 2) (n + 2)$ .

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

Step 1.3.6

Expand $(n + 2) (n + 2)$ using the FOIL Method.

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Step 1.3.6.1

Apply the distributive property.

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

Step 1.3.6.2

Apply the distributive property.

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

Step 1.3.6.3

Apply the distributive property.

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

Step 1.3.7

Simplify and combine like terms.

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Step 1.3.7.1

Simplify each term.

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Step 1.3.7.1.1

Multiply $n$ by $n$ .

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

Step 1.3.7.1.2

Move $2$ to the left of $n$ .

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

Step 1.3.7.1.3

Multiply $2$ by $2$ .

$lim_{n \to \infty} \frac{\frac{d}{d n} [n^{2} - 2 n + 1 - (n^{2} + 2 n + 2 n + 4)]}{\frac{d}{d n} [3 - n]}$

Step 1.3.7.2

Add $2 n$ and $2 n$ .

$lim_{n \to \infty} \frac{\frac{d}{d n} [n^{2} - 2 n + 1 - (n^{2} + 4 n + 4)]}{\frac{d}{d n} [3 - n]}$

Step 1.3.8

By the Sum Rule, the derivative of $n^{2} - 2 n + 1 - (n^{2} + 4 n + 4)$ with respect to $n$ is $\frac{d}{d n} [n^{2}] + \frac{d}{d n} [- 2 n] + \frac{d}{d n} [1] + \frac{d}{d n} [- (n^{2} + 4 n + 4)]$ .

$lim_{n \to \infty} \frac{\frac{d}{d n} [n^{2}] + \frac{d}{d n} [- 2 n] + \frac{d}{d n} [1] + \frac{d}{d n} [- (n^{2} + 4 n + 4)]}{\frac{d}{d n} [3 - n]}$

Step 1.3.9

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

$lim_{n \to \infty} \frac{2 n + \frac{d}{d n} [- 2 n] + \frac{d}{d n} [1] + \frac{d}{d n} [- (n^{2} + 4 n + 4)]}{\frac{d}{d n} [3 - n]}$

Step 1.3.10

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

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Step 1.3.10.1

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

$lim_{n \to \infty} \frac{2 n - 2 \frac{d}{d n} [n] + \frac{d}{d n} [1] + \frac{d}{d n} [- (n^{2} + 4 n + 4)]}{\frac{d}{d n} [3 - n]}$

Step 1.3.10.2

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

$lim_{n \to \infty} \frac{2 n - 2 \cdot 1 + \frac{d}{d n} [1] + \frac{d}{d n} [- (n^{2} + 4 n + 4)]}{\frac{d}{d n} [3 - n]}$

Step 1.3.10.3

Multiply $- 2$ by $1$ .

$lim_{n \to \infty} \frac{2 n - 2 + \frac{d}{d n} [1] + \frac{d}{d n} [- (n^{2} + 4 n + 4)]}{\frac{d}{d n} [3 - n]}$

Step 1.3.11

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

$lim_{n \to \infty} \frac{2 n - 2 + 0 + \frac{d}{d n} [- (n^{2} + 4 n + 4)]}{\frac{d}{d n} [3 - n]}$

Step 1.3.12

Evaluate $\frac{d}{d n} [- (n^{2} + 4 n + 4)]$ .

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Step 1.3.12.1

Since $- 1$ is constant with respect to $n$ , the derivative of $- (n^{2} + 4 n + 4)$ with respect to $n$ is $- \frac{d}{d n} [n^{2} + 4 n + 4]$ .

$lim_{n \to \infty} \frac{2 n - 2 + 0 - \frac{d}{d n} [n^{2} + 4 n + 4]}{\frac{d}{d n} [3 - n]}$

Step 1.3.12.2

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

$lim_{n \to \infty} \frac{2 n - 2 + 0 - (\frac{d}{d n} [n^{2}] + \frac{d}{d n} [4 n] + \frac{d}{d n} [4])}{\frac{d}{d n} [3 - n]}$

Step 1.3.12.3

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

$lim_{n \to \infty} \frac{2 n - 2 + 0 - (2 n + \frac{d}{d n} [4 n] + \frac{d}{d n} [4])}{\frac{d}{d n} [3 - n]}$

Step 1.3.12.4

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

$lim_{n \to \infty} \frac{2 n - 2 + 0 - (2 n + 4 \frac{d}{d n} [n] + \frac{d}{d n} [4])}{\frac{d}{d n} [3 - n]}$

Step 1.3.12.5

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

$lim_{n \to \infty} \frac{2 n - 2 + 0 - (2 n + 4 \cdot 1 + \frac{d}{d n} [4])}{\frac{d}{d n} [3 - n]}$

Step 1.3.12.6

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

$lim_{n \to \infty} \frac{2 n - 2 + 0 - (2 n + 4 \cdot 1 + 0)}{\frac{d}{d n} [3 - n]}$

Step 1.3.12.7

Multiply $4$ by $1$ .

$lim_{n \to \infty} \frac{2 n - 2 + 0 - (2 n + 4 + 0)}{\frac{d}{d n} [3 - n]}$

Step 1.3.12.8

Add $2 n + 4$ and $0$ .

$lim_{n \to \infty} \frac{2 n - 2 + 0 - (2 n + 4)}{\frac{d}{d n} [3 - n]}$

Step 1.3.13

Simplify.

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Step 1.3.13.1

Apply the distributive property.

$lim_{n \to \infty} \frac{2 n - 2 + 0 - (2 n) - 1 \cdot 4}{\frac{d}{d n} [3 - n]}$

Step 1.3.13.2

Combine terms.

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Step 1.3.13.2.1

Add $2 n - 2$ and $0$ .

$lim_{n \to \infty} \frac{2 n - 2 - (2 n) - 1 \cdot 4}{\frac{d}{d n} [3 - n]}$

Step 1.3.13.2.2

Multiply $2$ by $- 1$ .

$lim_{n \to \infty} \frac{2 n - 2 - 2 n - 1 \cdot 4}{\frac{d}{d n} [3 - n]}$

Step 1.3.13.2.3

Multiply $- 1$ by $4$ .

$lim_{n \to \infty} \frac{2 n - 2 - 2 n - 4}{\frac{d}{d n} [3 - n]}$

Step 1.3.13.2.4

Subtract $2 n$ from $2 n$ .

$lim_{n \to \infty} \frac{0 - 2 - 4}{\frac{d}{d n} [3 - n]}$

Step 1.3.13.2.5

Subtract $2$ from $0$ .

$lim_{n \to \infty} \frac{- 2 - 4}{\frac{d}{d n} [3 - n]}$

Step 1.3.13.2.6

Subtract $4$ from $- 2$ .

$lim_{n \to \infty} \frac{- 6}{\frac{d}{d n} [3 - n]}$

Step 1.3.14

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

$lim_{n \to \infty} \frac{- 6}{\frac{d}{d n} [3] + \frac{d}{d n} [- n]}$

Step 1.3.15

Since $3$ is constant with respect to $n$ , the derivative of $3$ with respect to $n$ is $0$ .

$lim_{n \to \infty} \frac{- 6}{0 + \frac{d}{d n} [- n]}$

Step 1.3.16

Evaluate $\frac{d}{d n} [- n]$ .

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Step 1.3.16.1

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

$lim_{n \to \infty} \frac{- 6}{0 - \frac{d}{d n} [n]}$

Step 1.3.16.2

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

$lim_{n \to \infty} \frac{- 6}{0 - 1 \cdot 1}$

Step 1.3.16.3

Multiply $- 1$ by $1$ .

$lim_{n \to \infty} \frac{- 6}{0 - 1}$

Step 1.3.17

Subtract $1$ from $0$ .

$lim_{n \to \infty} \frac{- 6}{- 1}$

Step 1.4

Move the negative one from the denominator of $\frac{- 6}{- 1}$ .

$lim_{n \to \infty} - 1 \cdot - 6$

Step 2

Evaluate the limit.

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Step 2.1

Evaluate the limit of $- 1 \cdot - 6$ which is constant as $n$ approaches $\infty$ .

$- 1 \cdot - 6$

Step 2.2

Multiply $- 1$ by $- 6$ .

$6$