Precalculus Examples

$lim_{x \to \frac{1}{2}} \frac{8 x^{3} - 1}{6 x^{2} - 5 x + 1}$

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.

$\frac{lim_{x \to \frac{1}{2}} 8 x^{3} - 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

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 $\frac{1}{2}$ .

$\frac{lim_{x \to \frac{1}{2}} 8 x^{3} - lim_{x \to \frac{1}{2}} 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.1.2

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

$\frac{8 lim_{x \to \frac{1}{2}} x^{3} - lim_{x \to \frac{1}{2}} 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.1.3

Move the exponent $3$ from $x^{3}$ outside the limit using the Limits Power Rule.

$\frac{8 {(lim_{x \to \frac{1}{2}} x)}^{3} - lim_{x \to \frac{1}{2}} 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.1.4

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{8 {(lim_{x \to \frac{1}{2}} x)}^{3} - 1 \cdot 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{8 {(\frac{1}{2})}^{3} - 1 \cdot 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

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

Apply the product rule to $\frac{1}{2}$ .

$\frac{8 \frac{1^{3}}{2^{3}} - 1 \cdot 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.3.1.2

One to any power is one.

$\frac{8 \frac{1}{2^{3}} - 1 \cdot 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.3.1.3

Raise $2$ to the power of $3$ .

$\frac{8 (\frac{1}{8}) - 1 \cdot 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.3.1.4

Cancel the common factor of $8$ .

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Step 1.1.2.3.1.4.1

Cancel the common factor.

$\frac{8 (\frac{1}{8}) - 1 \cdot 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.3.1.4.2

Rewrite the expression.

$\frac{1 - 1 \cdot 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.3.1.5

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to \frac{1}{2}} 6 x^{2} - 5 x + 1}$

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 $\frac{1}{2}$ .

$\frac{0}{lim_{x \to \frac{1}{2}} 6 x^{2} - lim_{x \to \frac{1}{2}} 5 x + lim_{x \to \frac{1}{2}} 1}$

Step 1.1.3.2

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{6 lim_{x \to \frac{1}{2}} x^{2} - lim_{x \to \frac{1}{2}} 5 x + lim_{x \to \frac{1}{2}} 1}$

Step 1.1.3.3

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

$\frac{0}{6 {(lim_{x \to \frac{1}{2}} x)}^{2} - lim_{x \to \frac{1}{2}} 5 x + lim_{x \to \frac{1}{2}} 1}$

Step 1.1.3.4

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{6 {(lim_{x \to \frac{1}{2}} x)}^{2} - 5 lim_{x \to \frac{1}{2}} x + lim_{x \to \frac{1}{2}} 1}$

Step 1.1.3.5

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{0}{6 {(lim_{x \to \frac{1}{2}} x)}^{2} - 5 lim_{x \to \frac{1}{2}} x + 1}$

Step 1.1.3.6

Evaluate the limits by plugging in $\frac{1}{2}$ for all occurrences of $x$ .

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Step 1.1.3.6.1

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{0}{6 {(\frac{1}{2})}^{2} - 5 lim_{x \to \frac{1}{2}} x + 1}$

Step 1.1.3.6.2

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{0}{6 {(\frac{1}{2})}^{2} - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7

Simplify the answer.

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Step 1.1.3.7.1

Simplify each term.

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Step 1.1.3.7.1.1

Apply the product rule to $\frac{1}{2}$ .

$\frac{0}{6 \frac{1^{2}}{2^{2}} - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7.1.2

One to any power is one.

$\frac{0}{6 \frac{1}{2^{2}} - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7.1.3

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

$\frac{0}{6 (\frac{1}{4}) - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7.1.4

Cancel the common factor of $2$ .

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Step 1.1.3.7.1.4.1

Factor $2$ out of $6$ .

$\frac{0}{2 (3) \frac{1}{4} - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7.1.4.2

Factor $2$ out of $4$ .

$\frac{0}{2 \cdot 3 \frac{1}{2 \cdot 2} - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7.1.4.3

Cancel the common factor.

$\frac{0}{2 \cdot 3 \frac{1}{2 \cdot 2} - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7.1.4.4

Rewrite the expression.

$\frac{0}{3 (\frac{1}{2}) - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7.1.5

Combine $3$ and $\frac{1}{2}$ .

$\frac{0}{\frac{3}{2} - 5 (\frac{1}{2}) + 1}$

Step 1.1.3.7.1.6

Combine $- 5$ and $\frac{1}{2}$ .

$\frac{0}{\frac{3}{2} + \frac{- 5}{2} + 1}$

Step 1.1.3.7.1.7

Move the negative in front of the fraction.

$\frac{0}{\frac{3}{2} - \frac{5}{2} + 1}$

Step 1.1.3.7.2

Combine the numerators over the common denominator.

$\frac{0}{1 + \frac{3 - 5}{2}}$

Step 1.1.3.7.3

Subtract $5$ from $3$ .

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

Step 1.1.3.7.4

Divide $- 2$ by $2$ .

$\frac{0}{1 - 1}$

Step 1.1.3.7.5

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.1.3.7.6

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

Undefined

$\frac{0}{0}$

Step 1.1.3.8

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.

$lim_{x \to \frac{1}{2}} \frac{8 x^{3} - 1}{6 x^{2} - 5 x + 1} = lim_{x \to \frac{1}{2}} \frac{\frac{d}{d x} [8 x^{3} - 1]}{\frac{d}{d x} [6 x^{2} - 5 x + 1]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to \frac{1}{2}} \frac{\frac{d}{d x} [8 x^{3} - 1]}{\frac{d}{d x} [6 x^{2} - 5 x + 1]}$

Step 1.3.2

By the Sum Rule, the derivative of $8 x^{3} - 1$ with respect to $x$ is $\frac{d}{d x} [8 x^{3}] + \frac{d}{d x} [- 1]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{d}{d x} [8 x^{3}] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [6 x^{2} - 5 x + 1]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [8 x^{3}]$ .

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Step 1.3.3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{3}$ with respect to $x$ is $8 \frac{d}{d x} [x^{3}]$ .

$lim_{x \to \frac{1}{2}} \frac{8 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [6 x^{2} - 5 x + 1]}$

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 $n = 3$ .

$lim_{x \to \frac{1}{2}} \frac{8 (3 x^{2}) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [6 x^{2} - 5 x + 1]}$

Step 1.3.3.3

Multiply $3$ by $8$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2} + \frac{d}{d x} [- 1]}{\frac{d}{d x} [6 x^{2} - 5 x + 1]}$

Step 1.3.4

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

$lim_{x \to \frac{1}{2}} \frac{24 x^{2} + 0}{\frac{d}{d x} [6 x^{2} - 5 x + 1]}$

Step 1.3.5

Add $24 x^{2}$ and $0$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{\frac{d}{d x} [6 x^{2} - 5 x + 1]}$

Step 1.3.6

By the Sum Rule, the derivative of $6 x^{2} - 5 x + 1$ with respect to $x$ is $\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [1]$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [1]}$

Step 1.3.7

Evaluate $\frac{d}{d x} [6 x^{2}]$ .

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Step 1.3.7.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{6 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [1]}$

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 $n = 2$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{6 (2 x) + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [1]}$

Step 1.3.7.3

Multiply $2$ by $6$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{12 x + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [1]}$

Step 1.3.8

Evaluate $\frac{d}{d x} [- 5 x]$ .

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Step 1.3.8.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{12 x - 5 \frac{d}{d x} [x] + \frac{d}{d x} [1]}$

Step 1.3.8.2

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

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{12 x - 5 \cdot 1 + \frac{d}{d x} [1]}$

Step 1.3.8.3

Multiply $- 5$ by $1$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{12 x - 5 + \frac{d}{d x} [1]}$

Step 1.3.9

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

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{12 x - 5 + 0}$

Step 1.3.10

Add $12 x - 5$ and $0$ .

$lim_{x \to \frac{1}{2}} \frac{24 x^{2}}{12 x - 5}$

Step 2

Evaluate the limit.

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Step 2.1

Move the term $24$ outside of the limit because it is constant with respect to $x$ .

$24 lim_{x \to \frac{1}{2}} \frac{x^{2}}{12 x - 5}$

Step 2.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$24 \frac{lim_{x \to \frac{1}{2}} x^{2}}{lim_{x \to \frac{1}{2}} 12 x - 5}$

Step 2.3

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

$24 \frac{{(lim_{x \to \frac{1}{2}} x)}^{2}}{lim_{x \to \frac{1}{2}} 12 x - 5}$

Step 2.4

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$24 \frac{{(lim_{x \to \frac{1}{2}} x)}^{2}}{lim_{x \to \frac{1}{2}} 12 x - lim_{x \to \frac{1}{2}} 5}$

Step 2.5

Move the term $12$ outside of the limit because it is constant with respect to $x$ .

$24 \frac{{(lim_{x \to \frac{1}{2}} x)}^{2}}{12 lim_{x \to \frac{1}{2}} x - lim_{x \to \frac{1}{2}} 5}$

Step 2.6

Evaluate the limit of $5$ which is constant as $x$ approaches $\frac{1}{2}$ .

$24 \frac{{(lim_{x \to \frac{1}{2}} x)}^{2}}{12 lim_{x \to \frac{1}{2}} x - 1 \cdot 5}$

Step 3

Evaluate the limits by plugging in $\frac{1}{2}$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$24 \frac{{(\frac{1}{2})}^{2}}{12 lim_{x \to \frac{1}{2}} x - 1 \cdot 5}$

Step 3.2

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$24 \frac{{(\frac{1}{2})}^{2}}{12 (\frac{1}{2}) - 1 \cdot 5}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Apply the product rule to $\frac{1}{2}$ .

$24 \frac{\frac{1^{2}}{2^{2}}}{12 (\frac{1}{2}) - 1 \cdot 5}$

Step 4.1.2

One to any power is one.

$24 \frac{\frac{1}{2^{2}}}{12 (\frac{1}{2}) - 1 \cdot 5}$

Step 4.1.3

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

$24 \frac{\frac{1}{4}}{12 (\frac{1}{2}) - 1 \cdot 5}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Factor $2$ out of $12$ .

$24 \frac{\frac{1}{4}}{2 (6) \frac{1}{2} - 1 \cdot 5}$

Step 4.2.1.2

Cancel the common factor.

$24 \frac{\frac{1}{4}}{2 \cdot 6 \frac{1}{2} - 1 \cdot 5}$

Step 4.2.1.3

Rewrite the expression.

$24 \frac{\frac{1}{4}}{6 - 1 \cdot 5}$

Step 4.2.2

Multiply $- 1$ by $5$ .

$24 \frac{\frac{1}{4}}{6 - 5}$

Step 4.2.3

Subtract $5$ from $6$ .

$24 \frac{\frac{1}{4}}{1}$

Step 4.3

Divide $\frac{1}{4}$ by $1$ .

$24 (\frac{1}{4})$

Step 4.4

Cancel the common factor of $4$ .

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Step 4.4.1

Factor $4$ out of $24$ .

$4 (6) \frac{1}{4}$

Step 4.4.2

Cancel the common factor.

$4 \cdot 6 \frac{1}{4}$

Step 4.4.3

Rewrite the expression.

$6$