Calculus Examples

$lim_{x \to \infty} \frac{\sqrt{x^{2} - 9}}{2 x - 6}$

Step 1

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

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Step 1.1

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

$\frac{lim_{x \to \infty} \sqrt{x^{2} - 9}}{lim_{x \to \infty} 2 x - 6}$

Step 1.2

As $x$ approaches $\infty$ for radicals, the value goes to $\infty$ .

$\frac{\infty}{lim_{x \to \infty} 2 x - 6}$

Step 1.3

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\frac{\infty}{\infty}$

Step 1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 2

Since $\frac{\infty}{\infty}$ 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.

$lim_{x \to \infty} \frac{\sqrt{x^{2} - 9}}{2 x - 6} = lim_{x \to \infty} \frac{\frac{d}{d x} [\sqrt{x^{2} - 9}]}{\frac{d}{d x} [2 x - 6]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [\sqrt{x^{2} - 9}]}{\frac{d}{d x} [2 x - 6]}$

Step 3.2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} - 9}$ as ${(x^{2} - 9)}^{\frac{1}{2}}$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [{(x^{2} - 9)}^{\frac{1}{2}}]}{\frac{d}{d x} [2 x - 6]}$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{2} - 9$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $x^{2} - 9$ .

$lim_{x \to \infty} \frac{\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.3.2

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

$lim_{x \to \infty} \frac{\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.3.3

Replace all occurrences of $u$ with $x^{2} - 9$ .

$lim_{x \to \infty} \frac{\frac{1}{2} {(x^{2} - 9)}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.4

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

$lim_{x \to \infty} \frac{\frac{1}{2} {(x^{2} - 9)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.5

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

$lim_{x \to \infty} \frac{\frac{1}{2} {(x^{2} - 9)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.6

Combine the numerators over the common denominator.

$lim_{x \to \infty} \frac{\frac{1}{2} {(x^{2} - 9)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.7

Simplify the numerator.

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Step 3.7.1

Multiply $- 1$ by $2$ .

$lim_{x \to \infty} \frac{\frac{1}{2} {(x^{2} - 9)}^{\frac{1 - 2}{2}} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.7.2

Subtract $2$ from $1$ .

$lim_{x \to \infty} \frac{\frac{1}{2} {(x^{2} - 9)}^{\frac{- 1}{2}} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.8

Move the negative in front of the fraction.

$lim_{x \to \infty} \frac{\frac{1}{2} {(x^{2} - 9)}^{- \frac{1}{2}} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.9

Combine $\frac{1}{2}$ and ${(x^{2} - 9)}^{- \frac{1}{2}}$ .

$lim_{x \to \infty} \frac{\frac{{(x^{2} - 9)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.10

Move ${(x^{2} - 9)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to \infty} \frac{\frac{1}{2 {(x^{2} - 9)}^{\frac{1}{2}}} \frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [2 x - 6]}$

Step 3.11

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

$lim_{x \to \infty} \frac{\frac{1}{2 {(x^{2} - 9)}^{\frac{1}{2}}} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9])}{\frac{d}{d x} [2 x - 6]}$

Step 3.12

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

$lim_{x \to \infty} \frac{\frac{1}{2 {(x^{2} - 9)}^{\frac{1}{2}}} (2 x + \frac{d}{d x} [- 9])}{\frac{d}{d x} [2 x - 6]}$

Step 3.13

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

$lim_{x \to \infty} \frac{\frac{1}{2 {(x^{2} - 9)}^{\frac{1}{2}}} (2 x + 0)}{\frac{d}{d x} [2 x - 6]}$

Step 3.14

Add $2 x$ and $0$ .

$lim_{x \to \infty} \frac{\frac{1}{2 {(x^{2} - 9)}^{\frac{1}{2}}} (2 x)}{\frac{d}{d x} [2 x - 6]}$

Step 3.15

Combine $2$ and $\frac{1}{2 {(x^{2} - 9)}^{\frac{1}{2}}}$ .

$lim_{x \to \infty} \frac{\frac{2}{2 {(x^{2} - 9)}^{\frac{1}{2}}} x}{\frac{d}{d x} [2 x - 6]}$

Step 3.16

Combine $\frac{2}{2 {(x^{2} - 9)}^{\frac{1}{2}}}$ and $x$ .

$lim_{x \to \infty} \frac{\frac{2 x}{2 {(x^{2} - 9)}^{\frac{1}{2}}}}{\frac{d}{d x} [2 x - 6]}$

Step 3.17

Cancel the common factor.

$lim_{x \to \infty} \frac{\frac{2 x}{2 {(x^{2} - 9)}^{\frac{1}{2}}}}{\frac{d}{d x} [2 x - 6]}$

Step 3.18

Rewrite the expression.

$lim_{x \to \infty} \frac{\frac{x}{{(x^{2} - 9)}^{\frac{1}{2}}}}{\frac{d}{d x} [2 x - 6]}$

Step 3.19

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

$lim_{x \to \infty} \frac{\frac{x}{{(x^{2} - 9)}^{\frac{1}{2}}}}{\frac{d}{d x} [2 x] + \frac{d}{d x} [- 6]}$

Step 3.20

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

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Step 3.20.1

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

$lim_{x \to \infty} \frac{\frac{x}{{(x^{2} - 9)}^{\frac{1}{2}}}}{2 \frac{d}{d x} [x] + \frac{d}{d x} [- 6]}$

Step 3.20.2

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

$lim_{x \to \infty} \frac{\frac{x}{{(x^{2} - 9)}^{\frac{1}{2}}}}{2 \cdot 1 + \frac{d}{d x} [- 6]}$

Step 3.20.3

Multiply $2$ by $1$ .

$lim_{x \to \infty} \frac{\frac{x}{{(x^{2} - 9)}^{\frac{1}{2}}}}{2 + \frac{d}{d x} [- 6]}$

Step 3.21

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

$lim_{x \to \infty} \frac{\frac{x}{{(x^{2} - 9)}^{\frac{1}{2}}}}{2 + 0}$

Step 3.22

Add $2$ and $0$ .

$lim_{x \to \infty} \frac{\frac{x}{{(x^{2} - 9)}^{\frac{1}{2}}}}{2}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

Rewrite ${(x^{2} - 9)}^{\frac{1}{2}}$ as $\sqrt{x^{2} - 9}$ .

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

Step 6

Multiply $\frac{x}{\sqrt{x^{2} - 9}}$ by $\frac{1}{2}$ .

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

Step 7

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

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

Step 8

Simplify.

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Step 8.1

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

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

Step 8.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

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

Step 9

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x = \sqrt{x^{2}}$ .

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

Step 10

Evaluate the limit.

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Step 10.1

Cancel the common factor of $x$ .

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

Step 10.2

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

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

Step 10.3

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

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

Step 10.4

Move the limit under the radical sign.

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

Step 11

Apply L'Hospital's rule.

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Step 11.1

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

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Step 11.1.1

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

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

Step 11.1.2

Evaluate the limit of the numerator.

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Step 11.1.2.1

Apply the distributive property.

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

Step 11.1.2.2

Apply the distributive property.

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

Step 11.1.2.3

Apply the distributive property.

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

Step 11.1.2.4

Reorder $x$ and $- 3$ .

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

Step 11.1.2.5

Raise $x$ to the power of $1$ .

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

Step 11.1.2.6

Raise $x$ to the power of $1$ .

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

Step 11.1.2.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 11.1.2.8

Simplify by adding terms.

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Step 11.1.2.8.1

Add $1$ and $1$ .

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

Step 11.1.2.8.2

Multiply $3$ by $- 3$ .

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

Step 11.1.2.8.3

Add $- 3 x$ and $3 x$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{\frac{lim_{x \to \infty} x^{2} + 0 - 9}{lim_{x \to \infty} x^{2}}}}$

Step 11.1.2.8.4

Subtract $9$ from $0$ .

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

Step 11.1.2.9

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

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

Step 11.1.3

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

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

Step 11.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 11.2

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

Step 11.3

Find the derivative of the numerator and denominator.

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Step 11.3.1

Differentiate the numerator and denominator.

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{\frac{d}{d x} [(x + 3) (x - 3)]}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 3$ and $g (x) = x - 3$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{(x + 3) \frac{d}{d x} [x - 3] + (x - 3) \frac{d}{d x} [x + 3]}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.3

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

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{(x + 3) (\frac{d}{d x} [x] + \frac{d}{d x} [- 3]) + (x - 3) \frac{d}{d x} [x + 3]}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.4

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

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{(x + 3) (1 + \frac{d}{d x} [- 3]) + (x - 3) \frac{d}{d x} [x + 3]}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.5

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

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{(x + 3) (1 + 0) + (x - 3) \frac{d}{d x} [x + 3]}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.6

Add $1$ and $0$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{(x + 3) \cdot 1 + (x - 3) \frac{d}{d x} [x + 3]}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.7

Multiply $x + 3$ by $1$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{x + 3 + (x - 3) \frac{d}{d x} [x + 3]}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.8

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

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{x + 3 + (x - 3) (\frac{d}{d x} [x] + \frac{d}{d x} [3])}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.9

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

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{x + 3 + (x - 3) (1 + \frac{d}{d x} [3])}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.10

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

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{x + 3 + (x - 3) (1 + 0)}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.11

Add $1$ and $0$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{x + 3 + (x - 3) \cdot 1}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.12

Multiply $x - 3$ by $1$ .

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

Step 11.3.13

Add $x$ and $x$ .

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

Step 11.3.14

Subtract $3$ from $3$ .

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

Step 11.3.15

Add $2 x$ and $0$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to \infty} \frac{2 x}{\frac{d}{d x} [x^{2}]}}}$

Step 11.3.16

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

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

Step 11.4

Reduce.

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Step 11.4.1

Cancel the common factor of $2$ .

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Step 11.4.1.1

Cancel the common factor.

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

Step 11.4.1.2

Rewrite the expression.

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

Step 11.4.2

Cancel the common factor of $x$ .

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Step 11.4.2.1

Cancel the common factor.

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

Step 11.4.2.2

Rewrite the expression.

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

Step 12

Evaluate the limit.

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Step 12.1

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

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

Step 12.2

Simplify the answer.

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Step 12.2.1

Any root of $1$ is $1$ .

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

Step 12.2.2

Cancel the common factor of $1$ .

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Step 12.2.2.1

Cancel the common factor.

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

Step 12.2.2.2

Rewrite the expression.

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

Step 12.2.3

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

$\frac{1}{2}$