Calculus Examples

$lim_{x \to 5} \frac{x^{2} - 9 x + 20}{x - 5}$

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.

$\frac{lim_{x \to 5} x^{2} - 9 x + 20}{lim_{x \to 5} x - 5}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$\frac{lim_{x \to 5} x^{2} - lim_{x \to 5} 9 x + lim_{x \to 5} 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.2

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

$\frac{{(lim_{x \to 5} x)}^{2} - lim_{x \to 5} 9 x + lim_{x \to 5} 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.3

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{{(lim_{x \to 5} x)}^{2} - 9 lim_{x \to 5} x + lim_{x \to 5} 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.4

Evaluate the limit of $20$ which is constant as $x$ approaches $5$ .

$\frac{{(lim_{x \to 5} x)}^{2} - 9 lim_{x \to 5} x + 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.5

Evaluate the limits by plugging in $5$ for all occurrences of $x$ .

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Step 1.1.2.5.1

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$\frac{5^{2} - 9 lim_{x \to 5} x + 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.5.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$\frac{5^{2} - 9 \cdot 5 + 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.6

Simplify the answer.

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Step 1.1.2.6.1

Simplify each term.

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Step 1.1.2.6.1.1

Raise $5$ to the power of $2$ .

$\frac{25 - 9 \cdot 5 + 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.6.1.2

Multiply $- 9$ by $5$ .

$\frac{25 - 45 + 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.6.2

Subtract $45$ from $25$ .

$\frac{- 20 + 20}{lim_{x \to 5} x - 5}$

Step 1.1.2.6.3

Add $- 20$ and $20$ .

$\frac{0}{lim_{x \to 5} x - 5}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$\frac{0}{lim_{x \to 5} x - lim_{x \to 5} 5}$

Step 1.1.3.1.2

Evaluate the limit of $5$ which is constant as $x$ approaches $5$ .

$\frac{0}{lim_{x \to 5} x - 1 \cdot 5}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$\frac{0}{5 - 1 \cdot 5}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Multiply $- 1$ by $5$ .

$\frac{0}{5 - 5}$

Step 1.1.3.3.2

Subtract $5$ from $5$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.4

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.

$lim_{x \to 5} \frac{x^{2} - 9 x + 20}{x - 5} = lim_{x \to 5} \frac{\frac{d}{d x} [x^{2} - 9 x + 20]}{\frac{d}{d x} [x - 5]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 5} \frac{\frac{d}{d x} [x^{2} - 9 x + 20]}{\frac{d}{d x} [x - 5]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} - 9 x + 20$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [20]$ .

$lim_{x \to 5} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [20]}{\frac{d}{d x} [x - 5]}$

Step 1.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$lim_{x \to 5} \frac{2 x + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [20]}{\frac{d}{d x} [x - 5]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- 9 x]$ .

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Step 1.3.4.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$lim_{x \to 5} \frac{2 x - 9 \frac{d}{d x} [x] + \frac{d}{d x} [20]}{\frac{d}{d x} [x - 5]}$

Step 1.3.4.2

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

$lim_{x \to 5} \frac{2 x - 9 \cdot 1 + \frac{d}{d x} [20]}{\frac{d}{d x} [x - 5]}$

Step 1.3.4.3

Multiply $- 9$ by $1$ .

$lim_{x \to 5} \frac{2 x - 9 + \frac{d}{d x} [20]}{\frac{d}{d x} [x - 5]}$

Step 1.3.5

Since $20$ is constant with respect to $x$ , the derivative of $20$ with respect to $x$ is $0$ .

$lim_{x \to 5} \frac{2 x - 9 + 0}{\frac{d}{d x} [x - 5]}$

Step 1.3.6

Add $2 x - 9$ and $0$ .

$lim_{x \to 5} \frac{2 x - 9}{\frac{d}{d x} [x - 5]}$

Step 1.3.7

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

$lim_{x \to 5} \frac{2 x - 9}{\frac{d}{d x} [x] + \frac{d}{d x} [- 5]}$

Step 1.3.8

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

$lim_{x \to 5} \frac{2 x - 9}{1 + \frac{d}{d x} [- 5]}$

Step 1.3.9

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$lim_{x \to 5} \frac{2 x - 9}{1 + 0}$

Step 1.3.10

Add $1$ and $0$ .

$lim_{x \to 5} \frac{2 x - 9}{1}$

Step 1.4

Divide $2 x - 9$ by $1$ .

$lim_{x \to 5} 2 x - 9$

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 $5$ .

$lim_{x \to 5} 2 x - lim_{x \to 5} 9$

Step 2.2

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

$2 lim_{x \to 5} x - lim_{x \to 5} 9$

Step 2.3

Evaluate the limit of $9$ which is constant as $x$ approaches $5$ .

$2 lim_{x \to 5} x - 1 \cdot 9$

Step 3

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$2 \cdot 5 - 1 \cdot 9$

Step 4

Simplify the answer.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $2$ by $5$ .

$10 - 1 \cdot 9$

Step 4.1.2

Multiply $- 1$ by $9$ .

$10 - 9$

Step 4.2

Subtract $9$ from $10$ .

$1$