Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 1.1.2.1.2

Move the exponent $2$ from ${(2 x + 3)}^{2}$ outside the limit using the Limits Power Rule.

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

Step 1.1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 1.1.2.1.4

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

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

Step 1.1.2.1.5

Evaluate the limit of $3$ which is constant as $x$ approaches $- 1$ .

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

Step 1.1.2.1.6

Evaluate the limit of $1$ which is constant as $x$ approaches $- 1$ .

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

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Multiply $2$ by $- 1$ .

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

Step 1.1.2.3.1.2

Add $- 2$ and $3$ .

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

Step 1.1.2.3.1.3

One to any power is one.

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

Step 1.1.2.3.1.4

Multiply $- 1$ by $1$ .

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

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to - 1} x^{2} + x}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{0}{lim_{x \to - 1} x^{2} + lim_{x \to - 1} x}$

Step 1.1.3.2

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

$\frac{0}{{(lim_{x \to - 1} x)}^{2} + lim_{x \to - 1} x}$

Step 1.1.3.3

Evaluate the limits by plugging in $- 1$ for all occurrences of $x$ .

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Step 1.1.3.3.1

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{0}{{(- 1)}^{2} + lim_{x \to - 1} x}$

Step 1.1.3.3.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{0}{{(- 1)}^{2} - 1}$

Step 1.1.3.4

Simplify the answer.

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Step 1.1.3.4.1

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

$\frac{0}{1 - 1}$

Step 1.1.3.4.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.1.3.4.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.5

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

Rewrite ${(2 x + 3)}^{2}$ as $(2 x + 3) (2 x + 3)$ .

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

Step 1.3.3

Expand $(2 x + 3) (2 x + 3)$ using the FOIL Method.

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Step 1.3.3.1

Apply the distributive property.

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

Step 1.3.3.2

Apply the distributive property.

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

Step 1.3.3.3

Apply the distributive property.

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

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

Rewrite using the commutative property of multiplication.

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

Step 1.3.4.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.3.4.1.2.1

Move $x$ .

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

Step 1.3.4.1.2.2

Multiply $x$ by $x$ .

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

Step 1.3.4.1.3

Multiply $2$ by $2$ .

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

Step 1.3.4.1.4

Multiply $3$ by $2$ .

$lim_{x \to - 1} \frac{\frac{d}{d x} [4 x^{2} + 6 x + 3 (2 x) + 3 \cdot 3 - 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.4.1.5

Multiply $2$ by $3$ .

$lim_{x \to - 1} \frac{\frac{d}{d x} [4 x^{2} + 6 x + 6 x + 3 \cdot 3 - 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.4.1.6

Multiply $3$ by $3$ .

$lim_{x \to - 1} \frac{\frac{d}{d x} [4 x^{2} + 6 x + 6 x + 9 - 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.4.2

Add $6 x$ and $6 x$ .

$lim_{x \to - 1} \frac{\frac{d}{d x} [4 x^{2} + 12 x + 9 - 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.5

By the Sum Rule, the derivative of $4 x^{2} + 12 x + 9 - 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [12 x] + \frac{d}{d x} [9] + \frac{d}{d x} [- 1]$ .

$lim_{x \to - 1} \frac{\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [12 x] + \frac{d}{d x} [9] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.6

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

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Step 1.3.6.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to - 1} \frac{4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [12 x] + \frac{d}{d x} [9] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.6.2

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

$lim_{x \to - 1} \frac{4 (2 x) + \frac{d}{d x} [12 x] + \frac{d}{d x} [9] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.6.3

Multiply $2$ by $4$ .

$lim_{x \to - 1} \frac{8 x + \frac{d}{d x} [12 x] + \frac{d}{d x} [9] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.7

Evaluate $\frac{d}{d x} [12 x]$ .

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Step 1.3.7.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x$ with respect to $x$ is $12 \frac{d}{d x} [x]$ .

$lim_{x \to - 1} \frac{8 x + 12 \frac{d}{d x} [x] + \frac{d}{d x} [9] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.7.2

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

$lim_{x \to - 1} \frac{8 x + 12 \cdot 1 + \frac{d}{d x} [9] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.7.3

Multiply $12$ by $1$ .

$lim_{x \to - 1} \frac{8 x + 12 + \frac{d}{d x} [9] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.8

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

$lim_{x \to - 1} \frac{8 x + 12 + 0 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.9

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

$lim_{x \to - 1} \frac{8 x + 12 + 0 + 0}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.10

Combine terms.

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Step 1.3.10.1

Add $8 x + 12$ and $0$ .

$lim_{x \to - 1} \frac{8 x + 12 + 0}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.10.2

Add $8 x + 12$ and $0$ .

$lim_{x \to - 1} \frac{8 x + 12}{\frac{d}{d x} [x^{2} + x]}$

Step 1.3.11

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

$lim_{x \to - 1} \frac{8 x + 12}{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x]}$

Step 1.3.12

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

$lim_{x \to - 1} \frac{8 x + 12}{2 x + \frac{d}{d x} [x]}$

Step 1.3.13

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

$lim_{x \to - 1} \frac{8 x + 12}{2 x + 1}$

Step 2

Evaluate the limit.

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Step 2.1

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 1$ .

$\frac{lim_{x \to - 1} 8 x + 12}{lim_{x \to - 1} 2 x + 1}$

Step 2.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{lim_{x \to - 1} 8 x + lim_{x \to - 1} 12}{lim_{x \to - 1} 2 x + 1}$

Step 2.3

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$\frac{8 lim_{x \to - 1} x + lim_{x \to - 1} 12}{lim_{x \to - 1} 2 x + 1}$

Step 2.4

Evaluate the limit of $12$ which is constant as $x$ approaches $- 1$ .

$\frac{8 lim_{x \to - 1} x + 12}{lim_{x \to - 1} 2 x + 1}$

Step 2.5

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{8 lim_{x \to - 1} x + 12}{lim_{x \to - 1} 2 x + lim_{x \to - 1} 1}$

Step 2.6

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

$\frac{8 lim_{x \to - 1} x + 12}{2 lim_{x \to - 1} x + lim_{x \to - 1} 1}$

Step 2.7

Evaluate the limit of $1$ which is constant as $x$ approaches $- 1$ .

$\frac{8 lim_{x \to - 1} x + 12}{2 lim_{x \to - 1} x + 1}$

Step 3

Evaluate the limits by plugging in $- 1$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{8 \cdot - 1 + 12}{2 lim_{x \to - 1} x + 1}$

Step 3.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{8 \cdot - 1 + 12}{2 \cdot - 1 + 1}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $8$ by $- 1$ .

$\frac{- 8 + 12}{2 \cdot - 1 + 1}$

Step 4.1.2

Add $- 8$ and $12$ .

$\frac{4}{2 \cdot - 1 + 1}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $- 1$ .

$\frac{4}{- 2 + 1}$

Step 4.2.2

Add $- 2$ and $1$ .

$\frac{4}{- 1}$

Step 4.3

Divide $4$ by $- 1$ .

$- 4$