Calculus Examples

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

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.

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

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 $2$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

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

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

Step 1.1.2.4

Evaluate the limits by plugging in $2$ for all occurrences of $x$ .

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Step 1.1.2.4.1

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Step 1.1.2.4.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Step 1.1.2.5

Simplify the answer.

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Step 1.1.2.5.1

Simplify each term.

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Step 1.1.2.5.1.1

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

$\frac{4 - 2 \cdot 2}{lim_{x \to 2} x^{2} + x - 6}$

Step 1.1.2.5.1.2

Multiply $- 2$ by $2$ .

$\frac{4 - 4}{lim_{x \to 2} x^{2} + x - 6}$

Step 1.1.2.5.2

Subtract $4$ from $4$ .

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

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 $2$ .

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

Step 1.1.3.2

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

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

Step 1.1.3.3

Evaluate the limit of $6$ which is constant as $x$ approaches $2$ .

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

Step 1.1.3.4

Evaluate the limits by plugging in $2$ for all occurrences of $x$ .

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Step 1.1.3.4.1

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Step 1.1.3.4.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{0}{2^{2} + 2 - 1 \cdot 6}$

Step 1.1.3.5

Simplify the answer.

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Step 1.1.3.5.1

Simplify each term.

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Step 1.1.3.5.1.1

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

$\frac{0}{4 + 2 - 1 \cdot 6}$

Step 1.1.3.5.1.2

Multiply $- 1$ by $6$ .

$\frac{0}{4 + 2 - 6}$

Step 1.1.3.5.2

Add $4$ and $2$ .

$\frac{0}{6 - 6}$

Step 1.1.3.5.3

Subtract $6$ from $6$ .

$\frac{0}{0}$

Step 1.1.3.5.4

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

Undefined

$\frac{0}{0}$

Step 1.1.3.6

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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Step 1.3.4.1

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

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

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 $n = 1$ .

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

Step 1.3.4.3

Multiply $- 2$ by $1$ .

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

Step 1.3.5

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

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

Step 1.3.6

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

$lim_{x \to 2} \frac{2 x - 2}{2 x + \frac{d}{d x} [x] + \frac{d}{d x} [- 6]}$

Step 1.3.7

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

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

Step 1.3.8

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

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

Step 1.3.9

Add $2 x + 1$ and $0$ .

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

Step 2

Evaluate the limit.

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Step 2.1

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $2$ .

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

Step 2.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

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

Step 2.6

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

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

Step 2.7

Evaluate the limit of $1$ which is constant as $x$ approaches $2$ .

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

Step 3

Evaluate the limits by plugging in $2$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Step 3.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $2$ by $2$ .

$\frac{4 - 1 \cdot 2}{2 \cdot 2 + 1}$

Step 4.1.2

Multiply $- 1$ by $2$ .

$\frac{4 - 2}{2 \cdot 2 + 1}$

Step 4.1.3

Subtract $2$ from $4$ .

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

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $2$ .

$\frac{2}{4 + 1}$

Step 4.2.2

Add $4$ and $1$ .

$\frac{2}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{2}{5}$

Decimal Form:

$0.4$