Calculus Examples

$lim_{a \to - 2} \frac{2 a^{3} - 2 a^{2} - 4 a + 16}{a + 2}$

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.

$\frac{lim_{a \to - 2} 2 a^{3} - 2 a^{2} - 4 a + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $a$ approaches $- 2$ .

$\frac{lim_{a \to - 2} 2 a^{3} - lim_{a \to - 2} 2 a^{2} - lim_{a \to - 2} 4 a + lim_{a \to - 2} 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.2

Move the term $2$ outside of the limit because it is constant with respect to $a$ .

$\frac{2 lim_{a \to - 2} a^{3} - lim_{a \to - 2} 2 a^{2} - lim_{a \to - 2} 4 a + lim_{a \to - 2} 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.3

Move the exponent $3$ from $a^{3}$ outside the limit using the Limits Power Rule.

$\frac{2 {(lim_{a \to - 2} a)}^{3} - lim_{a \to - 2} 2 a^{2} - lim_{a \to - 2} 4 a + lim_{a \to - 2} 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.4

Move the term $2$ outside of the limit because it is constant with respect to $a$ .

$\frac{2 {(lim_{a \to - 2} a)}^{3} - 2 lim_{a \to - 2} a^{2} - lim_{a \to - 2} 4 a + lim_{a \to - 2} 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.5

Move the exponent $2$ from $a^{2}$ outside the limit using the Limits Power Rule.

$\frac{2 {(lim_{a \to - 2} a)}^{3} - 2 {(lim_{a \to - 2} a)}^{2} - lim_{a \to - 2} 4 a + lim_{a \to - 2} 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.6

Move the term $4$ outside of the limit because it is constant with respect to $a$ .

$\frac{2 {(lim_{a \to - 2} a)}^{3} - 2 {(lim_{a \to - 2} a)}^{2} - 4 lim_{a \to - 2} a + lim_{a \to - 2} 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.7

Evaluate the limit of $16$ which is constant as $a$ approaches $- 2$ .

$\frac{2 {(lim_{a \to - 2} a)}^{3} - 2 {(lim_{a \to - 2} a)}^{2} - 4 lim_{a \to - 2} a + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.8

Evaluate the limits by plugging in $- 2$ for all occurrences of $a$ .

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Step 1.1.2.8.1

Evaluate the limit of $a$ by plugging in $- 2$ for $a$ .

$\frac{2 {(- 2)}^{3} - 2 {(lim_{a \to - 2} a)}^{2} - 4 lim_{a \to - 2} a + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.8.2

Evaluate the limit of $a$ by plugging in $- 2$ for $a$ .

$\frac{2 {(- 2)}^{3} - 2 {(- 2)}^{2} - 4 lim_{a \to - 2} a + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.8.3

Evaluate the limit of $a$ by plugging in $- 2$ for $a$ .

$\frac{2 {(- 2)}^{3} - 2 {(- 2)}^{2} - 4 \cdot - 2 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9

Simplify the answer.

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Step 1.1.2.9.1

Simplify each term.

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Step 1.1.2.9.1.1

Raise $- 2$ to the power of $3$ .

$\frac{2 \cdot - 8 - 2 {(- 2)}^{2} - 4 \cdot - 2 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.1.2

Multiply $2$ by $- 8$ .

$\frac{- 16 - 2 {(- 2)}^{2} - 4 \cdot - 2 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.1.3

Multiply $- 2$ by ${(- 2)}^{2}$ by adding the exponents.

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Step 1.1.2.9.1.3.1

Multiply $- 2$ by ${(- 2)}^{2}$ .

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Step 1.1.2.9.1.3.1.1

Raise $- 2$ to the power of $1$ .

$\frac{- 16 + {(- 2)}^{1} {(- 2)}^{2} - 4 \cdot - 2 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.1.3.1.2

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

$\frac{- 16 + {(- 2)}^{1 + 2} - 4 \cdot - 2 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.1.3.2

Add $1$ and $2$ .

$\frac{- 16 + {(- 2)}^{3} - 4 \cdot - 2 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.1.4

Raise $- 2$ to the power of $3$ .

$\frac{- 16 - 8 - 4 \cdot - 2 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.1.5

Multiply $- 4$ by $- 2$ .

$\frac{- 16 - 8 + 8 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.2

Subtract $8$ from $- 16$ .

$\frac{- 24 + 8 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.3

Add $- 24$ and $8$ .

$\frac{- 16 + 16}{lim_{a \to - 2} a + 2}$

Step 1.1.2.9.4

Add $- 16$ and $16$ .

$\frac{0}{lim_{a \to - 2} a + 2}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $a$ approaches $- 2$ .

$\frac{0}{lim_{a \to - 2} a + lim_{a \to - 2} 2}$

Step 1.1.3.1.2

Evaluate the limit of $2$ which is constant as $a$ approaches $- 2$ .

$\frac{0}{lim_{a \to - 2} a + 2}$

Step 1.1.3.2

Evaluate the limit of $a$ by plugging in $- 2$ for $a$ .

$\frac{0}{- 2 + 2}$

Step 1.1.3.3

Add $- 2$ and $2$ .

$\frac{0}{0}$

Step 1.1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.2

Since $\frac{0}{0}$ 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.

$lim_{a \to - 2} \frac{2 a^{3} - 2 a^{2} - 4 a + 16}{a + 2} = lim_{a \to - 2} \frac{\frac{d}{d a} [2 a^{3} - 2 a^{2} - 4 a + 16]}{\frac{d}{d a} [a + 2]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{a \to - 2} \frac{\frac{d}{d a} [2 a^{3} - 2 a^{2} - 4 a + 16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.2

By the Sum Rule, the derivative of $2 a^{3} - 2 a^{2} - 4 a + 16$ with respect to $a$ is $\frac{d}{d a} [2 a^{3}] + \frac{d}{d a} [- 2 a^{2}] + \frac{d}{d a} [- 4 a] + \frac{d}{d a} [16]$ .

$lim_{a \to - 2} \frac{\frac{d}{d a} [2 a^{3}] + \frac{d}{d a} [- 2 a^{2}] + \frac{d}{d a} [- 4 a] + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.3

Evaluate $\frac{d}{d a} [2 a^{3}]$ .

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Step 1.3.3.1

Since $2$ is constant with respect to $a$ , the derivative of $2 a^{3}$ with respect to $a$ is $2 \frac{d}{d a} [a^{3}]$ .

$lim_{a \to - 2} \frac{2 \frac{d}{d a} [a^{3}] + \frac{d}{d a} [- 2 a^{2}] + \frac{d}{d a} [- 4 a] + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d a} [a^{n}]$ is $n a^{n - 1}$ where $n = 3$ .

$lim_{a \to - 2} \frac{2 (3 a^{2}) + \frac{d}{d a} [- 2 a^{2}] + \frac{d}{d a} [- 4 a] + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.3.3

Multiply $3$ by $2$ .

$lim_{a \to - 2} \frac{6 a^{2} + \frac{d}{d a} [- 2 a^{2}] + \frac{d}{d a} [- 4 a] + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.4

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

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Step 1.3.4.1

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

$lim_{a \to - 2} \frac{6 a^{2} - 2 \frac{d}{d a} [a^{2}] + \frac{d}{d a} [- 4 a] + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.4.2

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

$lim_{a \to - 2} \frac{6 a^{2} - 2 (2 a) + \frac{d}{d a} [- 4 a] + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.4.3

Multiply $2$ by $- 2$ .

$lim_{a \to - 2} \frac{6 a^{2} - 4 a + \frac{d}{d a} [- 4 a] + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.5

Evaluate $\frac{d}{d a} [- 4 a]$ .

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Step 1.3.5.1

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

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4 \frac{d}{d a} [a] + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.5.2

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

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4 \cdot 1 + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.5.3

Multiply $- 4$ by $1$ .

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4 + \frac{d}{d a} [16]}{\frac{d}{d a} [a + 2]}$

Step 1.3.6

Since $16$ is constant with respect to $a$ , the derivative of $16$ with respect to $a$ is $0$ .

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4 + 0}{\frac{d}{d a} [a + 2]}$

Step 1.3.7

Add $6 a^{2} - 4 a - 4$ and $0$ .

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4}{\frac{d}{d a} [a + 2]}$

Step 1.3.8

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

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4}{\frac{d}{d a} [a] + \frac{d}{d a} [2]}$

Step 1.3.9

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

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4}{1 + \frac{d}{d a} [2]}$

Step 1.3.10

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

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4}{1 + 0}$

Step 1.3.11

Add $1$ and $0$ .

$lim_{a \to - 2} \frac{6 a^{2} - 4 a - 4}{1}$

Step 1.4

Divide $6 a^{2} - 4 a - 4$ by $1$ .

$lim_{a \to - 2} 6 a^{2} - 4 a - 4$

Step 2

Evaluate the limit.

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Step 2.1

Split the limit using the Sum of Limits Rule on the limit as $a$ approaches $- 2$ .

$lim_{a \to - 2} 6 a^{2} - lim_{a \to - 2} 4 a - lim_{a \to - 2} 4$

Step 2.2

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

$6 lim_{a \to - 2} a^{2} - lim_{a \to - 2} 4 a - lim_{a \to - 2} 4$

Step 2.3

Move the exponent $2$ from $a^{2}$ outside the limit using the Limits Power Rule.

$6 {(lim_{a \to - 2} a)}^{2} - lim_{a \to - 2} 4 a - lim_{a \to - 2} 4$

Step 2.4

Move the term $4$ outside of the limit because it is constant with respect to $a$ .

$6 {(lim_{a \to - 2} a)}^{2} - 4 lim_{a \to - 2} a - lim_{a \to - 2} 4$

Step 2.5

Evaluate the limit of $4$ which is constant as $a$ approaches $- 2$ .

$6 {(lim_{a \to - 2} a)}^{2} - 4 lim_{a \to - 2} a - 1 \cdot 4$

Step 3

Evaluate the limits by plugging in $- 2$ for all occurrences of $a$ .

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Step 3.1

Evaluate the limit of $a$ by plugging in $- 2$ for $a$ .

$6 {(- 2)}^{2} - 4 lim_{a \to - 2} a - 1 \cdot 4$

Step 3.2

Evaluate the limit of $a$ by plugging in $- 2$ for $a$ .

$6 {(- 2)}^{2} - 4 \cdot - 2 - 1 \cdot 4$

Step 4

Simplify the answer.

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Step 4.1

Simplify each term.

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Step 4.1.1

Raise $- 2$ to the power of $2$ .

$6 \cdot 4 - 4 \cdot - 2 - 1 \cdot 4$

Step 4.1.2

Multiply $6$ by $4$ .

$24 - 4 \cdot - 2 - 1 \cdot 4$

Step 4.1.3

Multiply $- 4$ by $- 2$ .

$24 + 8 - 1 \cdot 4$

Step 4.1.4

Multiply $- 1$ by $4$ .

$24 + 8 - 4$

Step 4.2

Add $24$ and $8$ .

$32 - 4$

Step 4.3

Subtract $4$ from $32$ .

$28$