Calculus Examples

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

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.

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

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 $0$ .

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

Step 1.1.2.1.2

Move the exponent $2$ from ${(2 x + 1)}^{2}$ outside the limit using the Limits Power Rule.

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

Step 1.1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

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

Step 1.1.2.1.4

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

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

Step 1.1.2.1.5

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

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

Step 1.1.2.1.6

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

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

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

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

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 $0$ .

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

Step 1.1.2.3.1.2

Add $0$ and $1$ .

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

Step 1.1.2.3.1.3

One to any power is one.

$\frac{1 - 1 \cdot 1}{lim_{x \to 0} x}$

Step 1.1.2.3.1.4

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to 0} x}$

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 0} x}$

Step 1.1.3

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

Rewrite ${(2 x + 1)}^{2}$ as $(2 x + 1) (2 x + 1)$ .

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

Step 1.3.3

Expand $(2 x + 1) (2 x + 1)$ using the FOIL Method.

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Step 1.3.3.1

Apply the distributive property.

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

Step 1.3.3.2

Apply the distributive property.

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

Step 1.3.3.3

Apply the distributive property.

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

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.

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

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 $x$ .

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

Step 1.3.4.1.2.2

Multiply $x$ by $x$ .

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

Step 1.3.4.1.3

Multiply $2$ by $2$ .

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

Step 1.3.4.1.4

Multiply $2$ by $1$ .

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

Step 1.3.4.1.5

Multiply $2 x$ by $1$ .

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

Step 1.3.4.1.6

Multiply $1$ by $1$ .

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

Step 1.3.4.2

Add $2 x$ and $2 x$ .

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

Step 1.3.5

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

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

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

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

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 $n = 2$ .

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

Step 1.3.6.3

Multiply $2$ by $4$ .

$lim_{x \to 0} \frac{8 x + \frac{d}{d x} [4 x] + \frac{d}{d x} [1] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 1.3.7

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

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Step 1.3.7.1

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

$lim_{x \to 0} \frac{8 x + 4 \frac{d}{d x} [x] + \frac{d}{d x} [1] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

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 $n = 1$ .

$lim_{x \to 0} \frac{8 x + 4 \cdot 1 + \frac{d}{d x} [1] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 1.3.7.3

Multiply $4$ by $1$ .

$lim_{x \to 0} \frac{8 x + 4 + \frac{d}{d x} [1] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 1.3.8

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

$lim_{x \to 0} \frac{8 x + 4 + 0 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 1.3.9

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

$lim_{x \to 0} \frac{8 x + 4 + 0 + 0}{\frac{d}{d x} [x]}$

Step 1.3.10

Combine terms.

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Step 1.3.10.1

Add $8 x + 4$ and $0$ .

$lim_{x \to 0} \frac{8 x + 4 + 0}{\frac{d}{d x} [x]}$

Step 1.3.10.2

Add $8 x + 4$ and $0$ .

$lim_{x \to 0} \frac{8 x + 4}{\frac{d}{d x} [x]}$

Step 1.3.11

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

$lim_{x \to 0} \frac{8 x + 4}{1}$

Step 1.4

Divide $8 x + 4$ by $1$ .

$lim_{x \to 0} 8 x + 4$

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 $x$ approaches $0$ .

$lim_{x \to 0} 8 x + lim_{x \to 0} 4$

Step 2.2

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

$8 lim_{x \to 0} x + lim_{x \to 0} 4$

Step 2.3

Evaluate the limit of $4$ which is constant as $x$ approaches $0$ .

$8 lim_{x \to 0} x + 4$

Step 3

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$8 \cdot 0 + 4$

Step 4

Simplify the answer.

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Step 4.1

Multiply $8$ by $0$ .

$0 + 4$

Step 4.2

Add $0$ and $4$ .

$4$