Calculus Examples

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

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.

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

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 $x$ approaches $- 1$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

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

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

Step 1.1.2.4

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.1.2.5

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

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

Step 1.1.2.6

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

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Step 1.1.2.6.1

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

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

Step 1.1.2.6.2

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

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

Step 1.1.2.7

Simplify the answer.

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Step 1.1.2.7.1

Simplify each term.

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Step 1.1.2.7.1.1

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

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

Step 1.1.2.7.1.2

Multiply $2$ by $1$ .

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

Step 1.1.2.7.1.3

Multiply $3$ by $- 1$ .

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

Step 1.1.2.7.2

Subtract $3$ from $2$ .

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

Step 1.1.2.7.3

Add $- 1$ and $1$ .

$\frac{0}{lim_{x \to - 1} x^{2} - 2 x - 3}$

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 $- 1$ .

$\frac{0}{lim_{x \to - 1} x^{2} - lim_{x \to - 1} 2 x - lim_{x \to - 1} 3}$

Step 1.1.3.2

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

$\frac{0}{{(lim_{x \to - 1} x)}^{2} - lim_{x \to - 1} 2 x - lim_{x \to - 1} 3}$

Step 1.1.3.3

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

$\frac{0}{{(lim_{x \to - 1} x)}^{2} - 2 lim_{x \to - 1} x - lim_{x \to - 1} 3}$

Step 1.1.3.4

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

$\frac{0}{{(lim_{x \to - 1} x)}^{2} - 2 lim_{x \to - 1} x - 1 \cdot 3}$

Step 1.1.3.5

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

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Step 1.1.3.5.1

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

$\frac{0}{{(- 1)}^{2} - 2 lim_{x \to - 1} x - 1 \cdot 3}$

Step 1.1.3.5.2

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

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

Step 1.1.3.6

Simplify the answer.

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Step 1.1.3.6.1

Simplify each term.

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Step 1.1.3.6.1.1

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

$\frac{0}{1 - 2 \cdot - 1 - 1 \cdot 3}$

Step 1.1.3.6.1.2

Multiply $- 2$ by $- 1$ .

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

Step 1.1.3.6.1.3

Multiply $- 1$ by $3$ .

$\frac{0}{1 + 2 - 3}$

Step 1.1.3.6.2

Add $1$ and $2$ .

$\frac{0}{3 - 3}$

Step 1.1.3.6.3

Subtract $3$ from $3$ .

$\frac{0}{0}$

Step 1.1.3.6.4

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

Undefined

$\frac{0}{0}$

Step 1.1.3.7

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

By the Sum Rule, the derivative of $2 x^{2} + 3 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$ .

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

Step 1.3.3

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

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Step 1.3.3.1

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

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

Step 1.3.3.2

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

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

Step 1.3.3.3

Multiply $2$ by $2$ .

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

Step 1.3.4

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 1.3.4.1

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

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

Step 1.3.4.2

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

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

Step 1.3.4.3

Multiply $3$ by $1$ .

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

Step 1.3.5

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

$lim_{x \to - 1} \frac{4 x + 3 + 0}{\frac{d}{d x} [x^{2} - 2 x - 3]}$

Step 1.3.6

Add $4 x + 3$ and $0$ .

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

Step 1.3.7

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

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

Step 1.3.8

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

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

Step 1.3.9

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

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Step 1.3.9.1

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

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

Step 1.3.9.2

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

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

Step 1.3.9.3

Multiply $- 2$ by $1$ .

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

Step 1.3.10

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

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

Step 1.3.11

Add $2 x - 2$ and $0$ .

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

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 $- 1$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

Evaluate the limit of $2$ which is constant as $x$ approaches $- 1$ .

$\frac{4 lim_{x \to - 1} x + 3}{2 lim_{x \to - 1} x - 1 \cdot 2}$

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 $x$ .

$\frac{4 \cdot - 1 + 3}{2 lim_{x \to - 1} x - 1 \cdot 2}$

Step 3.2

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

$\frac{4 \cdot - 1 + 3}{2 \cdot - 1 - 1 \cdot 2}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $4$ by $- 1$ .

$\frac{- 4 + 3}{2 \cdot - 1 - 1 \cdot 2}$

Step 4.1.2

Add $- 4$ and $3$ .

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

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $- 1$ .

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

Step 4.2.2

Multiply $- 1$ by $2$ .

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

Step 4.2.3

Subtract $2$ from $- 2$ .

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

Step 4.3

Dividing two negative values results in a positive value.

$\frac{1}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{4}$

Decimal Form:

$0.25$