Calculus Examples

$f (x) = {\begin{cases} \frac{x^{2} - 9}{x + 3} & x \neq - 3 \\ - 6 & x = - 3 \end{cases}$

Step 1

Find the limit of $\frac{x^{2} - 9}{x + 3}$ as $x$ approaches $- 3$ .

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Step 1.1

Apply L'Hospital's rule.

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Step 1.1.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 1.1.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to - 3} x^{2} - 9}{lim_{x \to - 3} x + 3}$

Step 1.1.1.2

Evaluate the limit of the numerator.

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Step 1.1.1.2.1

Evaluate the limit.

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Step 1.1.1.2.1.1

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

$\frac{lim_{x \to - 3} x^{2} - lim_{x \to - 3} 9}{lim_{x \to - 3} x + 3}$

Step 1.1.1.2.1.2

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

$\frac{{(lim_{x \to - 3} x)}^{2} - lim_{x \to - 3} 9}{lim_{x \to - 3} x + 3}$

Step 1.1.1.2.1.3

Evaluate the limit of $9$ which is constant as $x$ approaches $- 3$ .

$\frac{{(lim_{x \to - 3} x)}^{2} - 1 \cdot 9}{lim_{x \to - 3} x + 3}$

Step 1.1.1.2.2

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

$\frac{{(- 3)}^{2} - 1 \cdot 9}{lim_{x \to - 3} x + 3}$

Step 1.1.1.2.3

Simplify the answer.

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Step 1.1.1.2.3.1

Simplify each term.

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Step 1.1.1.2.3.1.1

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

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

Step 1.1.1.2.3.1.2

Multiply $- 1$ by $9$ .

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

Step 1.1.1.2.3.2

Subtract $9$ from $9$ .

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

Step 1.1.1.3

Evaluate the limit of the denominator.

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Step 1.1.1.3.1

Evaluate the limit.

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Step 1.1.1.3.1.1

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

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

Step 1.1.1.3.1.2

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

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

Step 1.1.1.3.2

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

$\frac{0}{- 3 + 3}$

Step 1.1.1.3.3

Add $- 3$ and $3$ .

$\frac{0}{0}$

Step 1.1.1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.1.2

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.

$lim_{x \to - 3} \frac{x^{2} - 9}{x + 3} = lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [x + 3]}$

Step 1.1.3

Find the derivative of the numerator and denominator.

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Step 1.1.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [x + 3]}$

Step 1.1.3.2

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

$lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x + 3]}$

Step 1.1.3.3

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

$lim_{x \to - 3} \frac{2 x + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x + 3]}$

Step 1.1.3.4

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$lim_{x \to - 3} \frac{2 x + 0}{\frac{d}{d x} [x + 3]}$

Step 1.1.3.5

Add $2 x$ and $0$ .

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

Step 1.1.3.6

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

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

Step 1.1.3.7

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

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

Step 1.1.3.8

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

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

Step 1.1.3.9

Add $1$ and $0$ .

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

Step 1.1.4

Divide $2 x$ by $1$ .

$lim_{x \to - 3} 2 x$

Step 1.2

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

$2 lim_{x \to - 3} x$

Step 1.3

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

$2 \cdot - 3$

Step 1.4

Multiply $2$ by $- 3$ .

$- 6$

Step 2

Replace the variable $x$ with $- 3$ in the expression.

$- 6$

Step 3

Since the limit of $\frac{x^{2} - 9}{x + 3}$ as $x$ approaches $- 3$ is equal to the function value at $x = - 3$ , the function is continuous at $x = - 3$ .

Continuous

Step 4