Calculus Examples

$lim_{x \to 0} \frac{{(4 x + 1)}^{9} - 1}{3 x}$

Step 1

Move the term $\frac{1}{3}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{3} lim_{x \to 0} \frac{{(4 x + 1)}^{9} - 1}{x}$

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.

$\frac{1}{3} \cdot \frac{lim_{x \to 0} {(4 x + 1)}^{9} - 1}{lim_{x \to 0} x}$

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 $0$ .

$\frac{1}{3} \cdot \frac{lim_{x \to 0} {(4 x + 1)}^{9} - lim_{x \to 0} 1}{lim_{x \to 0} x}$

Step 2.1.2.1.2

Move the exponent $9$ from ${(4 x + 1)}^{9}$ outside the limit using the Limits Power Rule.

$\frac{1}{3} \cdot \frac{{(lim_{x \to 0} 4 x + 1)}^{9} - lim_{x \to 0} 1}{lim_{x \to 0} x}$

Step 2.1.2.1.3

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

$\frac{1}{3} \cdot \frac{{(lim_{x \to 0} 4 x + lim_{x \to 0} 1)}^{9} - lim_{x \to 0} 1}{lim_{x \to 0} x}$

Step 2.1.2.1.4

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

$\frac{1}{3} \cdot \frac{{(4 lim_{x \to 0} x + lim_{x \to 0} 1)}^{9} - lim_{x \to 0} 1}{lim_{x \to 0} x}$

Step 2.1.2.1.5

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

$\frac{1}{3} \cdot \frac{{(4 lim_{x \to 0} x + 1)}^{9} - lim_{x \to 0} 1}{lim_{x \to 0} x}$

Step 2.1.2.1.6

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

$\frac{1}{3} \cdot \frac{{(4 lim_{x \to 0} x + 1)}^{9} - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.2

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

$\frac{1}{3} \cdot \frac{{(4 \cdot 0 + 1)}^{9} - 1 \cdot 1}{lim_{x \to 0} x}$

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 $4$ by $0$ .

$\frac{1}{3} \cdot \frac{{(0 + 1)}^{9} - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.3.1.2

Add $0$ and $1$ .

$\frac{1}{3} \cdot \frac{1^{9} - 1 \cdot 1}{lim_{x \to 0} x}$

Step 2.1.2.3.1.3

One to any power is one.

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

Step 2.1.2.3.1.4

Multiply $- 1$ by $1$ .

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

Step 2.1.2.3.2

Subtract $1$ from $1$ .

$\frac{1}{3} \cdot \frac{0}{lim_{x \to 0} x}$

Step 2.1.3

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

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

Step 2.1.4

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

Undefined

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

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.

$lim_{x \to 0} \frac{{(4 x + 1)}^{9} - 1}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [{(4 x + 1)}^{9} - 1]}{\frac{d}{d x} [x]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$\frac{1}{3} lim_{x \to 0} \frac{\frac{d}{d x} [{(4 x + 1)}^{9} - 1]}{\frac{d}{d x} [x]}$

Step 2.3.2

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

$\frac{1}{3} lim_{x \to 0} \frac{\frac{d}{d x} [{(4 x + 1)}^{9}] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3

Evaluate $\frac{d}{d x} [{(4 x + 1)}^{9}]$ .

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Step 2.3.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{9}$ and $g (x) = 4 x + 1$ .

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Step 2.3.3.1.1

To apply the Chain Rule, set $u$ as $4 x + 1$ .

$\frac{1}{3} lim_{x \to 0} \frac{\frac{d}{d u} [u^{9}] \frac{d}{d x} [4 x + 1] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 9$ .

$\frac{1}{3} lim_{x \to 0} \frac{9 u^{8} \frac{d}{d x} [4 x + 1] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.1.3

Replace all occurrences of $u$ with $4 x + 1$ .

$\frac{1}{3} lim_{x \to 0} \frac{9 {(4 x + 1)}^{8} \frac{d}{d x} [4 x + 1] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.2

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

$\frac{1}{3} lim_{x \to 0} \frac{9 {(4 x + 1)}^{8} (\frac{d}{d x} [4 x] + \frac{d}{d x} [1]) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.3

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

$\frac{1}{3} lim_{x \to 0} \frac{9 {(4 x + 1)}^{8} (4 \frac{d}{d x} [x] + \frac{d}{d x} [1]) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.4

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

$\frac{1}{3} lim_{x \to 0} \frac{9 {(4 x + 1)}^{8} (4 \cdot 1 + \frac{d}{d x} [1]) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.5

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

$\frac{1}{3} lim_{x \to 0} \frac{9 {(4 x + 1)}^{8} (4 \cdot 1 + 0) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.6

Multiply $4$ by $1$ .

$\frac{1}{3} lim_{x \to 0} \frac{9 {(4 x + 1)}^{8} (4 + 0) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.7

Add $4$ and $0$ .

$\frac{1}{3} lim_{x \to 0} \frac{9 {(4 x + 1)}^{8} \cdot 4 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.3.8

Multiply $4$ by $9$ .

$\frac{1}{3} lim_{x \to 0} \frac{36 {(4 x + 1)}^{8} + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x]}$

Step 2.3.4

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

$\frac{1}{3} lim_{x \to 0} \frac{36 {(4 x + 1)}^{8} + 0}{\frac{d}{d x} [x]}$

Step 2.3.5

Add $36 {(4 x + 1)}^{8}$ and $0$ .

$\frac{1}{3} lim_{x \to 0} \frac{36 {(4 x + 1)}^{8}}{\frac{d}{d x} [x]}$

Step 2.3.6

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

$\frac{1}{3} lim_{x \to 0} \frac{36 {(4 x + 1)}^{8}}{1}$

Step 2.4

Divide $36 {(4 x + 1)}^{8}$ by $1$ .

$\frac{1}{3} lim_{x \to 0} 36 {(4 x + 1)}^{8}$

Step 3

Evaluate the limit.

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Step 3.1

Move the term $36$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{3} \cdot 36 lim_{x \to 0} {(4 x + 1)}^{8}$

Step 3.2

Move the exponent $8$ from ${(4 x + 1)}^{8}$ outside the limit using the Limits Power Rule.

$\frac{1}{3} \cdot 36 {(lim_{x \to 0} 4 x + 1)}^{8}$

Step 3.3

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

$\frac{1}{3} \cdot 36 {(lim_{x \to 0} 4 x + lim_{x \to 0} 1)}^{8}$

Step 3.4

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

$\frac{1}{3} \cdot 36 {(4 lim_{x \to 0} x + lim_{x \to 0} 1)}^{8}$

Step 3.5

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

$\frac{1}{3} \cdot 36 {(4 lim_{x \to 0} x + 1)}^{8}$

Step 4

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

$\frac{1}{3} \cdot 36 {(4 \cdot 0 + 1)}^{8}$

Step 5

Simplify the answer.

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Step 5.1

Cancel the common factor of $3$ .

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Step 5.1.1

Factor $3$ out of $36$ .

$\frac{1}{3} \cdot (3 (12)) {(4 \cdot 0 + 1)}^{8}$

Step 5.1.2

Cancel the common factor.

$\frac{1}{3} \cdot (3 \cdot 12) {(4 \cdot 0 + 1)}^{8}$

Step 5.1.3

Rewrite the expression.

$12 {(4 \cdot 0 + 1)}^{8}$

Step 5.2

Multiply $4$ by $0$ .

$12 {(0 + 1)}^{8}$

Step 5.3

Add $0$ and $1$ .

$12 \cdot 1^{8}$

Step 5.4

One to any power is one.

$12 \cdot 1$

Step 5.5

Multiply $12$ by $1$ .

$12$