Calculus Examples

$lim_{x \to \infty} \frac{x^{5}}{e^{4 x}}$

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.

$\frac{lim_{x \to \infty} x^{5}}{lim_{x \to \infty} e^{4 x}}$

Step 1.2

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

$\frac{\infty}{lim_{x \to \infty} e^{4 x}}$

Step 1.3

Since the exponent $4 x$ approaches $\infty$ , the quantity $e^{4 x}$ approaches $\infty$ .

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

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.

$lim_{x \to \infty} \frac{x^{5}}{e^{4 x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x^{5}]}{\frac{d}{d x} [e^{4 x}]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [x^{5}]}{\frac{d}{d x} [e^{4 x}]}$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$lim_{x \to \infty} \frac{5 x^{4}}{\frac{d}{d x} [e^{4 x}]}$

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 $f (x) = e^{x}$ and $g (x) = 4 x$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $4 x$ .

$lim_{x \to \infty} \frac{5 x^{4}}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [4 x]}$

Step 3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$lim_{x \to \infty} \frac{5 x^{4}}{e^{u} \frac{d}{d x} [4 x]}$

Step 3.3.3

Replace all occurrences of $u$ with $4 x$ .

$lim_{x \to \infty} \frac{5 x^{4}}{e^{4 x} \frac{d}{d x} [4 x]}$

Step 3.4

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{5 x^{4}}{e^{4 x} (4 \frac{d}{d x} [x])}$

Step 3.5

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

$lim_{x \to \infty} \frac{5 x^{4}}{e^{4 x} (4 \cdot 1)}$

Step 3.6

Multiply $4$ by $1$ .

$lim_{x \to \infty} \frac{5 x^{4}}{e^{4 x} \cdot 4}$

Step 3.7

Move $4$ to the left of $e^{4 x}$ .

$lim_{x \to \infty} \frac{5 x^{4}}{4 \cdot e^{4 x}}$

Step 3.8

Multiply $4$ by $e^{4 x}$ .

$lim_{x \to \infty} \frac{5 x^{4}}{4 e^{4 x}}$

Step 4

Move the term $\frac{5}{4}$ outside of the limit because it is constant with respect to $x$ .

$\frac{5}{4} lim_{x \to \infty} \frac{x^{4}}{e^{4 x}}$

Step 5

Apply L'Hospital's rule.

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Step 5.1

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

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Step 5.1.1

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

$\frac{5}{4} \cdot \frac{lim_{x \to \infty} x^{4}}{lim_{x \to \infty} e^{4 x}}$

Step 5.1.2

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

$\frac{5}{4} \cdot \frac{\infty}{lim_{x \to \infty} e^{4 x}}$

Step 5.1.3

Since the exponent $4 x$ approaches $\infty$ , the quantity $e^{4 x}$ approaches $\infty$ .

$\frac{5}{4} \cdot \frac{\infty}{\infty}$

Step 5.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{5}{4} \cdot \frac{\infty}{\infty}$

Step 5.2

$lim_{x \to \infty} \frac{x^{4}}{e^{4 x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x^{4}]}{\frac{d}{d x} [e^{4 x}]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$\frac{5}{4} lim_{x \to \infty} \frac{\frac{d}{d x} [x^{4}]}{\frac{d}{d x} [e^{4 x}]}$

Step 5.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{\frac{d}{d x} [e^{4 x}]}$

Step 5.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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Step 5.3.3.1

To apply the Chain Rule, set $u$ as $4 x$ .

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [4 x]}$

Step 5.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{e^{u} \frac{d}{d x} [4 x]}$

Step 5.3.3.3

Replace all occurrences of $u$ with $4 x$ .

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{e^{4 x} \frac{d}{d x} [4 x]}$

Step 5.3.4

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{e^{4 x} \cdot 4 \frac{d}{d x} [x]}$

Step 5.3.5

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

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{e^{4 x} \cdot 4 \cdot 1}$

Step 5.3.6

Multiply $4$ by $1$ .

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{e^{4 x} \cdot 4}$

Step 5.3.7

Move $4$ to the left of $e^{4 x}$ .

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{4 e^{4 x}}$

Step 5.4

Cancel the common factor of $4$ .

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Step 5.4.1

Cancel the common factor.

$\frac{5}{4} lim_{x \to \infty} \frac{4 x^{3}}{4 e^{4 x}}$

Step 5.4.2

Rewrite the expression.

$\frac{5}{4} lim_{x \to \infty} \frac{x^{3}}{e^{4 x}}$

Step 6

Apply L'Hospital's rule.

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Step 6.1

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

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Step 6.1.1

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

$\frac{5}{4} \cdot \frac{lim_{x \to \infty} x^{3}}{lim_{x \to \infty} e^{4 x}}$

Step 6.1.2

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

$\frac{5}{4} \cdot \frac{\infty}{lim_{x \to \infty} e^{4 x}}$

Step 6.1.3

Since the exponent $4 x$ approaches $\infty$ , the quantity $e^{4 x}$ approaches $\infty$ .

$\frac{5}{4} \cdot \frac{\infty}{\infty}$

Step 6.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{5}{4} \cdot \frac{\infty}{\infty}$

Step 6.2

$lim_{x \to \infty} \frac{x^{3}}{e^{4 x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x^{3}]}{\frac{d}{d x} [e^{4 x}]}$

Step 6.3

Find the derivative of the numerator and denominator.

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Step 6.3.1

Differentiate the numerator and denominator.

$\frac{5}{4} lim_{x \to \infty} \frac{\frac{d}{d x} [x^{3}]}{\frac{d}{d x} [e^{4 x}]}$

Step 6.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$\frac{5}{4} lim_{x \to \infty} \frac{3 x^{2}}{\frac{d}{d x} [e^{4 x}]}$

Step 6.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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Step 6.3.3.1

To apply the Chain Rule, set $u$ as $4 x$ .

$\frac{5}{4} lim_{x \to \infty} \frac{3 x^{2}}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [4 x]}$

Step 6.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{5}{4} lim_{x \to \infty} \frac{3 x^{2}}{e^{u} \frac{d}{d x} [4 x]}$

Step 6.3.3.3

Replace all occurrences of $u$ with $4 x$ .

$\frac{5}{4} lim_{x \to \infty} \frac{3 x^{2}}{e^{4 x} \frac{d}{d x} [4 x]}$

Step 6.3.4

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$\frac{5}{4} lim_{x \to \infty} \frac{3 x^{2}}{e^{4 x} \cdot 4 \frac{d}{d x} [x]}$

Step 6.3.5

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

$\frac{5}{4} lim_{x \to \infty} \frac{3 x^{2}}{e^{4 x} \cdot 4 \cdot 1}$

Step 6.3.6

Multiply $4$ by $1$ .

$\frac{5}{4} lim_{x \to \infty} \frac{3 x^{2}}{e^{4 x} \cdot 4}$

Step 6.3.7

Move $4$ to the left of $e^{4 x}$ .

$\frac{5}{4} lim_{x \to \infty} \frac{3 x^{2}}{4 e^{4 x}}$

Step 7

Move the term $\frac{3}{4}$ outside of the limit because it is constant with respect to $x$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{x^{2}}{e^{4 x}}$

Step 8

Apply L'Hospital's rule.

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Step 8.1

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

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Step 8.1.1

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

$\frac{5}{4} \cdot \frac{3}{4} \frac{lim_{x \to \infty} x^{2}}{lim_{x \to \infty} e^{4 x}}$

Step 8.1.2

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

$\frac{5}{4} \cdot \frac{3}{4} \frac{\infty}{lim_{x \to \infty} e^{4 x}}$

Step 8.1.3

Since the exponent $4 x$ approaches $\infty$ , the quantity $e^{4 x}$ approaches $\infty$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{\infty}{\infty}$

Step 8.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{5}{4} \cdot \frac{3}{4} \frac{\infty}{\infty}$

Step 8.2

$lim_{x \to \infty} \frac{x^{2}}{e^{4 x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{4 x}]}$

Step 8.3

Find the derivative of the numerator and denominator.

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Step 8.3.1

Differentiate the numerator and denominator.

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{\frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{4 x}]}$

Step 8.3.2

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

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{\frac{d}{d x} [e^{4 x}]}$

Step 8.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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Step 8.3.3.1

To apply the Chain Rule, set $u$ as $4 x$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [4 x]}$

Step 8.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{e^{u} \frac{d}{d x} [4 x]}$

Step 8.3.3.3

Replace all occurrences of $u$ with $4 x$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{e^{4 x} \frac{d}{d x} [4 x]}$

Step 8.3.4

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{e^{4 x} \cdot 4 \frac{d}{d x} [x]}$

Step 8.3.5

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

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{e^{4 x} \cdot 4 \cdot 1}$

Step 8.3.6

Multiply $4$ by $1$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{e^{4 x} \cdot 4}$

Step 8.3.7

Move $4$ to the left of $e^{4 x}$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{4 e^{4 x}}$

Step 8.4

Cancel the common factor of $2$ and $4$ .

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Step 8.4.1

Factor $2$ out of $2 x$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 (x)}{4 e^{4 x}}$

Step 8.4.2

Cancel the common factors.

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Step 8.4.2.1

Factor $2$ out of $4 e^{4 x}$ .

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 (x)}{2 \cdot 2 e^{4 x}}$

Step 8.4.2.2

Cancel the common factor.

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{2 x}{2 \cdot 2 e^{4 x}}$

Step 8.4.2.3

Rewrite the expression.

$\frac{5}{4} \cdot \frac{3}{4} lim_{x \to \infty} \frac{x}{2 e^{4 x}}$

Step 9

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

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{x}{e^{4 x}}$

Step 10

Apply L'Hospital's rule.

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Step 10.1

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

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Step 10.1.1

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

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} \frac{lim_{x \to \infty} x}{lim_{x \to \infty} e^{4 x}}$

Step 10.1.2

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

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} \frac{\infty}{lim_{x \to \infty} e^{4 x}}$

Step 10.1.3

Since the exponent $4 x$ approaches $\infty$ , the quantity $e^{4 x}$ approaches $\infty$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} \frac{\infty}{\infty}$

Step 10.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} \frac{\infty}{\infty}$

Step 10.2

$lim_{x \to \infty} \frac{x}{e^{4 x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [e^{4 x}]}$

Step 10.3

Find the derivative of the numerator and denominator.

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Step 10.3.1

Differentiate the numerator and denominator.

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [e^{4 x}]}$

Step 10.3.2

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

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{1}{\frac{d}{d x} [e^{4 x}]}$

Step 10.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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Step 10.3.3.1

To apply the Chain Rule, set $u$ as $4 x$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{1}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [4 x]}$

Step 10.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{1}{e^{u} \frac{d}{d x} [4 x]}$

Step 10.3.3.3

Replace all occurrences of $u$ with $4 x$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{1}{e^{4 x} \frac{d}{d x} [4 x]}$

Step 10.3.4

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{1}{e^{4 x} \cdot 4 \frac{d}{d x} [x]}$

Step 10.3.5

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

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{1}{e^{4 x} \cdot 4 \cdot 1}$

Step 10.3.6

Multiply $4$ by $1$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{1}{e^{4 x} \cdot 4}$

Step 10.3.7

Move $4$ to the left of $e^{4 x}$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} lim_{x \to \infty} \frac{1}{4 e^{4 x}}$

Step 11

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

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} \frac{1}{4} lim_{x \to \infty} \frac{1}{e^{4 x}}$

Step 12

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{e^{4 x}}$ approaches $0$ .

$\frac{5}{4} \cdot \frac{3}{4} \frac{1}{2} \frac{1}{4} \cdot 0$

Step 13

Simplify the answer.

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Step 13.1

Multiply $\frac{5}{4} \cdot \frac{3}{4}$ .

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Step 13.1.1

Multiply $\frac{5}{4}$ by $\frac{3}{4}$ .

$\frac{5 \cdot 3}{4 \cdot 4} \cdot \frac{1}{2} \frac{1}{4} \cdot 0$

Step 13.1.2

Multiply $5$ by $3$ .

$\frac{15}{4 \cdot 4} \cdot \frac{1}{2} \frac{1}{4} \cdot 0$

Step 13.1.3

Multiply $4$ by $4$ .

$\frac{15}{16} \cdot \frac{1}{2} \frac{1}{4} \cdot 0$

Step 13.2

Multiply $\frac{15}{16} \cdot \frac{1}{2}$ .

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Step 13.2.1

Multiply $\frac{15}{16}$ by $\frac{1}{2}$ .

$\frac{15}{16 \cdot 2} \cdot \frac{1}{4} \cdot 0$

Step 13.2.2

Multiply $16$ by $2$ .

$\frac{15}{32} \cdot \frac{1}{4} \cdot 0$

Step 13.3

Multiply $\frac{15}{32} \cdot \frac{1}{4}$ .

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Step 13.3.1

Multiply $\frac{15}{32}$ by $\frac{1}{4}$ .

$\frac{15}{32 \cdot 4} \cdot 0$

Step 13.3.2

Multiply $32$ by $4$ .

$\frac{15}{128} \cdot 0$

Step 13.4

Multiply $\frac{15}{128}$ by $0$ .

$0$