Simplify limit as x approaches 9 of ( square root of x-3)/(x-9)

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

Apply L\'Hospital\'s rule.

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

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

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

Evaluate the limit of the numerator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

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

Move the limit under the radical sign.

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

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

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

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

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

Simplify the answer.

Simplify each term.

Rewrite $9$ as $3^{2}$ .

$\frac{\sqrt{3^{2}} - 1 \cdot 3}{lim_{x \to 9} x - 9}$

Pull terms out from under the radical, assuming positive real numbers.

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

Multiply $- 1$ by $3$ .

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

Subtract $3$ from $3$ .

$\frac{0}{lim_{x \to 9} x - 9}$

Evaluate the limit of the denominator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{0}{lim_{x \to 9} x - lim_{x \to 9} 9}$

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

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9}$

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

$\frac{0}{9 - 1 \cdot 9}$

Simplify the answer.

Multiply $- 1$ by $9$ .

$\frac{0}{9 - 9}$

Subtract $9$ from $9$ .

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} = lim_{x \to 9} \frac{\frac{d}{d x} [\sqrt{x} - 3]}{\frac{d}{d x} [x - 9]}$

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

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

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

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

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$lim_{x \to 9} \frac{\frac{d}{d x} [x^{\frac{1}{2}}] + \frac{d}{d x} [- 3]}{\frac{d}{d x} [x - 9]}$

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

$lim_{x \to 9} \frac{\frac{1}{2} x^{\frac{1}{2} - 1} + \frac{d}{d x} [- 3]}{\frac{d}{d x} [x - 9]}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to 9} \frac{\frac{1}{2} x^{\frac{1}{2} - 1 \cdot \frac{2}{2}} + \frac{d}{d x} [- 3]}{\frac{d}{d x} [x - 9]}$

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

$lim_{x \to 9} \frac{\frac{1}{2} x^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} + \frac{d}{d x} [- 3]}{\frac{d}{d x} [x - 9]}$

Combine the numerators over the common denominator.

$lim_{x \to 9} \frac{\frac{1}{2} x^{\frac{1 - 1 \cdot 2}{2}} + \frac{d}{d x} [- 3]}{\frac{d}{d x} [x - 9]}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$lim_{x \to 9} \frac{\frac{1}{2} x^{\frac{1 - 2}{2}} + \frac{d}{d x} [- 3]}{\frac{d}{d x} [x - 9]}$

Subtract $2$ from $1$ .

$lim_{x \to 9} \frac{\frac{1}{2} x^{\frac{- 1}{2}} + \frac{d}{d x} [- 3]}{\frac{d}{d x} [x - 9]}$

Move the negative in front of the fraction.

$lim_{x \to 9} \frac{\frac{1}{2} x^{- \frac{1}{2}} + \frac{d}{d x} [- 3]}{\frac{d}{d x} [x - 9]}$

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

$lim_{x \to 9} \frac{\frac{1}{2} x^{- \frac{1}{2}} + 0}{\frac{d}{d x} [x - 9]}$

Simplify.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to 9} \frac{\frac{1}{2} \cdot \frac{1}{x^{\frac{1}{2}}} + 0}{\frac{d}{d x} [x - 9]}$

Combine terms.

Multiply $\frac{1}{2}$ by $\frac{1}{x^{\frac{1}{2}}}$ .

$lim_{x \to 9} \frac{\frac{1}{2 x^{\frac{1}{2}}} + 0}{\frac{d}{d x} [x - 9]}$

Add $\frac{1}{2 x^{\frac{1}{2}}}$ and $0$ .

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

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

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

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

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

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

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

Add $1$ and $0$ .

$lim_{x \to 9} \frac{\frac{1}{2 x^{\frac{1}{2}}}}{1}$

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 9} \frac{1}{2 x^{\frac{1}{2}}} \cdot 1$

Rewrite $x^{\frac{1}{2}}$ as $\sqrt{x}$ .

$lim_{x \to 9} \frac{1}{2 \sqrt{x}} \cdot 1$

Multiply $\frac{1}{2 \sqrt{x}}$ by $1$ .

$lim_{x \to 9} \frac{1}{2 \sqrt{x}}$

Evaluate the limit.

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} lim_{x \to 9} \frac{1}{\sqrt{x}}$

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $9$ .

$\frac{1}{2} \cdot \frac{lim_{x \to 9} 1}{lim_{x \to 9} \sqrt{x}}$

Evaluate the limit of $1$ which is constant as $x$ approaches $9$ .

$\frac{1}{2} \cdot \frac{1}{lim_{x \to 9} \sqrt{x}}$

Move the limit under the radical sign.

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to 9} x}}$

Simplify the denominator.

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{9}}$

Rewrite $9$ as $3^{2}$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{3^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$\frac{1}{2} \cdot \frac{1}{3}$

Multiply $\frac{1}{2} \cdot \frac{1}{3}$ .

Multiply $\frac{1}{2}$ by $\frac{1}{3}$ .

$\frac{1}{2 \cdot 3}$

Multiply $2$ by $3$ .

$\frac{1}{6}$

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{6}$

Decimal Form:

$0.1\overline{6}$

Simplify limit as x approaches 9 of ( square root of x-3)/(x-9)

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