Calculus Examples

$lim_{x \to - 3} \frac{x^{2} + 3 x}{4 (x + 3)}$

Step 1

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

$\frac{1}{4} lim_{x \to - 3} \frac{x^{2} + 3 x}{x + 3}$

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.

$\frac{1}{4} \cdot \frac{lim_{x \to - 3} x^{2} + 3 x}{lim_{x \to - 3} x + 3}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{1}{4} \cdot \frac{lim_{x \to - 3} x^{2} + lim_{x \to - 3} 3 x}{lim_{x \to - 3} x + 3}$

Step 2.1.2.2

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

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

Step 2.1.2.3

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

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

Step 2.1.2.4

Evaluate the limits by plugging in $- 3$ for all occurrences of $x$ .

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Step 2.1.2.4.1

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

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

Step 2.1.2.4.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

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

Step 2.1.2.5

Simplify the answer.

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Step 2.1.2.5.1

Simplify each term.

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Step 2.1.2.5.1.1

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

$\frac{1}{4} \cdot \frac{9 + 3 \cdot - 3}{lim_{x \to - 3} x + 3}$

Step 2.1.2.5.1.2

Multiply $3$ by $- 3$ .

$\frac{1}{4} \cdot \frac{9 - 9}{lim_{x \to - 3} x + 3}$

Step 2.1.2.5.2

Subtract $9$ from $9$ .

$\frac{1}{4} \cdot \frac{0}{lim_{x \to - 3} x + 3}$

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

Evaluate the limit.

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Step 2.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{1}{4} \cdot \frac{0}{lim_{x \to - 3} x + lim_{x \to - 3} 3}$

Step 2.1.3.1.2

Evaluate the limit of $3$ which is constant as $x$ approaches $- 3$ .

$\frac{1}{4} \cdot \frac{0}{lim_{x \to - 3} x + 3}$

Step 2.1.3.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{1}{4} \cdot \frac{0}{- 3 + 3}$

Step 2.1.3.3

Add $- 3$ and $3$ .

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

Step 2.1.3.4

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

Undefined

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

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

Step 2.3.2

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

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

Step 2.3.3

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

$\frac{1}{4} lim_{x \to - 3} \frac{2 x + \frac{d}{d x} [3 x]}{\frac{d}{d x} [x + 3]}$

Step 2.3.4

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

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Step 2.3.4.1

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

$\frac{1}{4} lim_{x \to - 3} \frac{2 x + 3 \frac{d}{d x} [x]}{\frac{d}{d x} [x + 3]}$

Step 2.3.4.2

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

$\frac{1}{4} lim_{x \to - 3} \frac{2 x + 3 \cdot 1}{\frac{d}{d x} [x + 3]}$

Step 2.3.4.3

Multiply $3$ by $1$ .

$\frac{1}{4} lim_{x \to - 3} \frac{2 x + 3}{\frac{d}{d x} [x + 3]}$

Step 2.3.5

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

$\frac{1}{4} lim_{x \to - 3} \frac{2 x + 3}{\frac{d}{d x} [x] + \frac{d}{d x} [3]}$

Step 2.3.6

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

$\frac{1}{4} lim_{x \to - 3} \frac{2 x + 3}{1 + \frac{d}{d x} [3]}$

Step 2.3.7

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$\frac{1}{4} lim_{x \to - 3} \frac{2 x + 3}{1 + 0}$

Step 2.3.8

Add $1$ and $0$ .

$\frac{1}{4} lim_{x \to - 3} \frac{2 x + 3}{1}$

Step 2.4

Divide $2 x + 3$ by $1$ .

$\frac{1}{4} lim_{x \to - 3} 2 x + 3$

Step 3

Evaluate the limit.

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Step 3.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{1}{4} (lim_{x \to - 3} 2 x + lim_{x \to - 3} 3)$

Step 3.2

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

$\frac{1}{4} (2 lim_{x \to - 3} x + lim_{x \to - 3} 3)$

Step 3.3

Evaluate the limit of $3$ which is constant as $x$ approaches $- 3$ .

$\frac{1}{4} (2 lim_{x \to - 3} x + 3)$

Step 4

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

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

Step 5

Simplify the answer.

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Step 5.1

Multiply $2$ by $- 3$ .

$\frac{1}{4} (- 6 + 3)$

Step 5.2

Add $- 6$ and $3$ .

$\frac{1}{4} \cdot - 3$

Step 5.3

Combine $\frac{1}{4}$ and $- 3$ .

$\frac{- 3}{4}$

Step 5.4

Move the negative in front of the fraction.

$- \frac{3}{4}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$- \frac{3}{4}$

Decimal Form:

$- 0.75$