Calculus Examples

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

Step 1

Rewrite ${(1 + 2 x)}^{\frac{1}{3}}$ as $\sqrt[3]{1 + 2 x}$ .

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

Step 2

Apply L'Hospital's rule.

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Step 2.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 2.1.1

Take the limit of the numerator and the limit of the denominator.

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

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Evaluate the limit.

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Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

Move the limit under the radical sign.

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

Step 2.1.2.1.3

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

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

Step 2.1.2.1.4

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

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

Step 2.1.2.1.5

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

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

Step 2.1.2.1.6

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

$\frac{\sqrt[3]{1 + 2 lim_{x \to 0} x} - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.2

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

$\frac{\sqrt[3]{1 + 2 \cdot 0} - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.3

Simplify the answer.

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Step 2.1.2.3.1

Simplify each term.

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Step 2.1.2.3.1.1

Multiply $2$ by $0$ .

$\frac{\sqrt[3]{1 + 0} - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.3.1.2

Add $1$ and $0$ .

$\frac{\sqrt[3]{1} - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.3.1.3

Any root of $1$ is $1$ .

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

Step 2.1.2.3.1.4

Multiply $- 1$ by $1$ .

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

Step 2.1.2.3.2

Subtract $1$ from $1$ .

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

Step 2.1.3

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

$\frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{0}{0}$

Step 2.2

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

Step 2.3.2

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

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

Step 2.3.3

Evaluate $\frac{d}{d x} [\sqrt[3]{1 + 2 x}]$ .

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Step 2.3.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{1 + 2 x}$ as ${(1 + 2 x)}^{\frac{1}{3}}$ .

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

Step 2.3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{3}}$ and $g (x) = 1 + 2 x$ .

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Step 2.3.3.2.1

To apply the Chain Rule, set $u$ as $1 + 2 x$ .

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

Step 2.3.3.2.2

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

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

Step 2.3.3.2.3

Replace all occurrences of $u$ with $1 + 2 x$ .

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

Step 2.3.3.3

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

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

Step 2.3.3.4

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

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

Step 2.3.3.5

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

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

Step 2.3.3.6

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

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

Step 2.3.3.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 2.3.3.8

Combine $- 1$ and $\frac{3}{3}$ .

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

Step 2.3.3.9

Combine the numerators over the common denominator.

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

Step 2.3.3.10

Simplify the numerator.

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Step 2.3.3.10.1

Multiply $- 1$ by $3$ .

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

Step 2.3.3.10.2

Subtract $3$ from $1$ .

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

Step 2.3.3.11

Move the negative in front of the fraction.

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

Step 2.3.3.12

Multiply $2$ by $1$ .

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

Step 2.3.3.13

Add $0$ and $2$ .

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

Step 2.3.3.14

Combine $\frac{1}{3}$ and ${(1 + 2 x)}^{- \frac{2}{3}}$ .

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

Step 2.3.3.15

Combine $\frac{{(1 + 2 x)}^{- \frac{2}{3}}}{3}$ and $2$ .

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

Step 2.3.3.16

Move $2$ to the left of ${(1 + 2 x)}^{- \frac{2}{3}}$ .

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

Step 2.3.3.17

Move ${(1 + 2 x)}^{- \frac{2}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.3.4

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

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

Step 2.3.5

Add $\frac{2}{3 {(1 + 2 x)}^{\frac{2}{3}}}$ and $0$ .

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

Step 2.3.6

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

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

Step 2.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.5

Multiply $\frac{2}{3 {(1 + 2 x)}^{\frac{2}{3}}}$ by $1$ .

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

Step 3

Evaluate the limit.

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Step 3.1

Move the term $\frac{2}{3}$ outside of the limit because it is constant with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Move the exponent $\frac{2}{3}$ from ${(1 + 2 x)}^{\frac{2}{3}}$ outside the limit using the Limits Power Rule.

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

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

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

Step 4

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

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

Step 5

Simplify the answer.

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Step 5.1

Combine.

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

Step 5.2

Multiply $2$ by $1$ .

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

Step 5.3

Simplify the denominator.

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Step 5.3.1

Multiply $2$ by $0$ .

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

Step 5.3.2

Add $1$ and $0$ .

$\frac{2}{3 \cdot 1^{\frac{2}{3}}}$

Step 5.3.3

One to any power is one.

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

Step 5.4

Multiply $3$ by $1$ .

$\frac{2}{3}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{2}{3}$

Decimal Form:

$0.\overline{6}$