Calculus Examples

$lim_{x \to 0} \frac{{(2 - 3 x)}^{7} - 128}{4 x}$

Step 1

Move the term $\frac{1}{4}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{4} lim_{x \to 0} \frac{{(2 - 3 x)}^{7} - 128}{x}$

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.

$\frac{1}{4} \cdot \frac{lim_{x \to 0} {(2 - 3 x)}^{7} - 128}{lim_{x \to 0} x}$

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 $0$ .

$\frac{1}{4} \cdot \frac{lim_{x \to 0} {(2 - 3 x)}^{7} - lim_{x \to 0} 128}{lim_{x \to 0} x}$

Step 2.1.2.1.2

Move the exponent $7$ from ${(2 - 3 x)}^{7}$ outside the limit using the Limits Power Rule.

$\frac{1}{4} \cdot \frac{{(lim_{x \to 0} 2 - 3 x)}^{7} - lim_{x \to 0} 128}{lim_{x \to 0} x}$

Step 2.1.2.1.3

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

$\frac{1}{4} \cdot \frac{{(lim_{x \to 0} 2 - lim_{x \to 0} 3 x)}^{7} - lim_{x \to 0} 128}{lim_{x \to 0} x}$

Step 2.1.2.1.4

Evaluate the limit of $2$ which is constant as $x$ approaches $0$ .

$\frac{1}{4} \cdot \frac{{(2 - lim_{x \to 0} 3 x)}^{7} - lim_{x \to 0} 128}{lim_{x \to 0} x}$

Step 2.1.2.1.5

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

$\frac{1}{4} \cdot \frac{{(2 - 3 lim_{x \to 0} x)}^{7} - lim_{x \to 0} 128}{lim_{x \to 0} x}$

Step 2.1.2.1.6

Evaluate the limit of $128$ which is constant as $x$ approaches $0$ .

$\frac{1}{4} \cdot \frac{{(2 - 3 lim_{x \to 0} x)}^{7} - 1 \cdot 128}{lim_{x \to 0} x}$

Step 2.1.2.2

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

$\frac{1}{4} \cdot \frac{{(2 - 3 \cdot 0)}^{7} - 1 \cdot 128}{lim_{x \to 0} x}$

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 $- 3$ by $0$ .

$\frac{1}{4} \cdot \frac{{(2 + 0)}^{7} - 1 \cdot 128}{lim_{x \to 0} x}$

Step 2.1.2.3.1.2

Add $2$ and $0$ .

$\frac{1}{4} \cdot \frac{2^{7} - 1 \cdot 128}{lim_{x \to 0} x}$

Step 2.1.2.3.1.3

Raise $2$ to the power of $7$ .

$\frac{1}{4} \cdot \frac{128 - 1 \cdot 128}{lim_{x \to 0} x}$

Step 2.1.2.3.1.4

Multiply $- 1$ by $128$ .

$\frac{1}{4} \cdot \frac{128 - 128}{lim_{x \to 0} x}$

Step 2.1.2.3.2

Subtract $128$ from $128$ .

$\frac{1}{4} \cdot \frac{0}{lim_{x \to 0} x}$

Step 2.1.3

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

$\frac{1}{4} \cdot \frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{1}{4} \cdot \frac{0}{0}$

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.

$lim_{x \to 0} \frac{{(2 - 3 x)}^{7} - 128}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [{(2 - 3 x)}^{7} - 128]}{\frac{d}{d x} [x]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$\frac{1}{4} lim_{x \to 0} \frac{\frac{d}{d x} [{(2 - 3 x)}^{7} - 128]}{\frac{d}{d x} [x]}$

Step 2.3.2

By the Sum Rule, the derivative of ${(2 - 3 x)}^{7} - 128$ with respect to $x$ is $\frac{d}{d x} [{(2 - 3 x)}^{7}] + \frac{d}{d x} [- 128]$ .

$\frac{1}{4} lim_{x \to 0} \frac{\frac{d}{d x} [{(2 - 3 x)}^{7}] + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3

Evaluate $\frac{d}{d x} [{(2 - 3 x)}^{7}]$ .

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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 $f (x) = x^{7}$ and $g (x) = 2 - 3 x$ .

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Step 2.3.3.1.1

To apply the Chain Rule, set $u$ as $2 - 3 x$ .

$\frac{1}{4} lim_{x \to 0} \frac{\frac{d}{d u} [u^{7}] \frac{d}{d x} [2 - 3 x] + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

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 $n = 7$ .

$\frac{1}{4} lim_{x \to 0} \frac{7 u^{6} \frac{d}{d x} [2 - 3 x] + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3.1.3

Replace all occurrences of $u$ with $2 - 3 x$ .

$\frac{1}{4} lim_{x \to 0} \frac{7 {(2 - 3 x)}^{6} \frac{d}{d x} [2 - 3 x] + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3.2

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

$\frac{1}{4} lim_{x \to 0} \frac{7 {(2 - 3 x)}^{6} (\frac{d}{d x} [2] + \frac{d}{d x} [- 3 x]) + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3.3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{1}{4} lim_{x \to 0} \frac{7 {(2 - 3 x)}^{6} (0 + \frac{d}{d x} [- 3 x]) + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3.4

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

$\frac{1}{4} lim_{x \to 0} \frac{7 {(2 - 3 x)}^{6} (0 - 3 \frac{d}{d x} [x]) + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3.5

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

$\frac{1}{4} lim_{x \to 0} \frac{7 {(2 - 3 x)}^{6} (0 - 3 \cdot 1) + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3.6

Multiply $- 3$ by $1$ .

$\frac{1}{4} lim_{x \to 0} \frac{7 {(2 - 3 x)}^{6} (0 - 3) + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3.7

Subtract $3$ from $0$ .

$\frac{1}{4} lim_{x \to 0} \frac{7 {(2 - 3 x)}^{6} \cdot - 3 + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.3.8

Multiply $- 3$ by $7$ .

$\frac{1}{4} lim_{x \to 0} \frac{- 21 {(2 - 3 x)}^{6} + \frac{d}{d x} [- 128]}{\frac{d}{d x} [x]}$

Step 2.3.4

Since $- 128$ is constant with respect to $x$ , the derivative of $- 128$ with respect to $x$ is $0$ .

$\frac{1}{4} lim_{x \to 0} \frac{- 21 {(2 - 3 x)}^{6} + 0}{\frac{d}{d x} [x]}$

Step 2.3.5

Add $- 21 {(2 - 3 x)}^{6}$ and $0$ .

$\frac{1}{4} lim_{x \to 0} \frac{- 21 {(2 - 3 x)}^{6}}{\frac{d}{d x} [x]}$

Step 2.3.6

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

$\frac{1}{4} lim_{x \to 0} \frac{- 21 {(2 - 3 x)}^{6}}{1}$

Step 2.4

Divide $- 21 {(2 - 3 x)}^{6}$ by $1$ .

$\frac{1}{4} lim_{x \to 0} - 21 {(2 - 3 x)}^{6}$

Step 3

Evaluate the limit.

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Step 3.1

Move the term $- 21$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{4} \cdot - 21 lim_{x \to 0} {(2 - 3 x)}^{6}$

Step 3.2

Move the exponent $6$ from ${(2 - 3 x)}^{6}$ outside the limit using the Limits Power Rule.

$\frac{1}{4} \cdot - 21 {(lim_{x \to 0} 2 - 3 x)}^{6}$

Step 3.3

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

$\frac{1}{4} \cdot - 21 {(lim_{x \to 0} 2 - lim_{x \to 0} 3 x)}^{6}$

Step 3.4

Evaluate the limit of $2$ which is constant as $x$ approaches $0$ .

$\frac{1}{4} \cdot - 21 {(2 - lim_{x \to 0} 3 x)}^{6}$

Step 3.5

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

$\frac{1}{4} \cdot - 21 {(2 - 3 lim_{x \to 0} x)}^{6}$

Step 4

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

$\frac{1}{4} \cdot - 21 {(2 - 3 \cdot 0)}^{6}$

Step 5

Simplify the answer.

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Step 5.1

Combine $\frac{1}{4}$ and $- 21$ .

$\frac{- 21}{4} {(2 - 3 \cdot 0)}^{6}$

Step 5.2

Move the negative in front of the fraction.

$- \frac{21}{4} {(2 - 3 \cdot 0)}^{6}$

Step 5.3

Multiply $- 3$ by $0$ .

$- \frac{21}{4} {(2 + 0)}^{6}$

Step 5.4

Add $2$ and $0$ .

$- \frac{21}{4} \cdot 2^{6}$

Step 5.5

Raise $2$ to the power of $6$ .

$- \frac{21}{4} \cdot 64$

Step 5.6

Cancel the common factor of $4$ .

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Step 5.6.1

Move the leading negative in $- \frac{21}{4}$ into the numerator.

$\frac{- 21}{4} \cdot 64$

Step 5.6.2

Factor $4$ out of $64$ .

$\frac{- 21}{4} \cdot (4 (16))$

Step 5.6.3

Cancel the common factor.

$\frac{- 21}{4} \cdot (4 \cdot 16)$

Step 5.6.4

Rewrite the expression.

$- 21 \cdot 16$

Step 5.7

Multiply $- 21$ by $16$ .

$- 336$