Calculus Examples

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

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.

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

Step 1.2

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

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

Step 1.3

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ 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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

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 $g (x) = 5 x$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $5 x$ .

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

Step 3.3.2

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

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

Step 3.3.3

Replace all occurrences of $u$ with $5 x$ .

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

Step 3.4

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

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

Step 3.5

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

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

Step 3.6

Multiply $5$ by $1$ .

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

Step 3.7

Move $5$ to the left of $e^{5 x}$ .

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

Step 3.8

Multiply $5$ by $e^{5 x}$ .

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

Step 4

Cancel the common factor of $5$ .

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Step 4.1

Cancel the common factor.

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

Step 4.2

Rewrite the expression.

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

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.

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

Step 5.1.2

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

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

Step 5.1.3

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ approaches $\infty$ .

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

Step 5.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 5.2

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

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

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

Step 5.3.2

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

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

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 $g (x) = 5 x$ .

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Step 5.3.3.1

To apply the Chain Rule, set $u$ as $5 x$ .

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

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 $a$ = $e$ .

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

Step 5.3.3.3

Replace all occurrences of $u$ with $5 x$ .

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

Step 5.3.4

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

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

Step 5.3.5

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

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

Step 5.3.6

Multiply $5$ by $1$ .

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

Step 5.3.7

Move $5$ to the left of $e^{5 x}$ .

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

Step 6

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

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

Step 7

Apply L'Hospital's rule.

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Step 7.1

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

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Step 7.1.1

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

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

Step 7.1.2

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

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

Step 7.1.3

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ approaches $\infty$ .

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

Step 7.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 7.2

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

Step 7.3

Find the derivative of the numerator and denominator.

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Step 7.3.1

Differentiate the numerator and denominator.

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

Step 7.3.2

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

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

Step 7.3.3

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 $g (x) = 5 x$ .

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Step 7.3.3.1

To apply the Chain Rule, set $u$ as $5 x$ .

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

Step 7.3.3.2

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

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

Step 7.3.3.3

Replace all occurrences of $u$ with $5 x$ .

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

Step 7.3.4

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

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

Step 7.3.5

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

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

Step 7.3.6

Multiply $5$ by $1$ .

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

Step 7.3.7

Move $5$ to the left of $e^{5 x}$ .

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

Step 8

Move the term $\frac{3}{5}$ outside of the limit because it is constant with respect to $x$ .

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

Step 9

Apply L'Hospital's rule.

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Step 9.1

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

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Step 9.1.1

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

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

Step 9.1.2

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

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

Step 9.1.3

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ approaches $\infty$ .

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

Step 9.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 9.2

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

Step 9.3

Find the derivative of the numerator and denominator.

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Step 9.3.1

Differentiate the numerator and denominator.

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

Step 9.3.2

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

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

Step 9.3.3

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 $g (x) = 5 x$ .

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Step 9.3.3.1

To apply the Chain Rule, set $u$ as $5 x$ .

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

Step 9.3.3.2

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

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

Step 9.3.3.3

Replace all occurrences of $u$ with $5 x$ .

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

Step 9.3.4

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

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

Step 9.3.5

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

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

Step 9.3.6

Multiply $5$ by $1$ .

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

Step 9.3.7

Move $5$ to the left of $e^{5 x}$ .

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

Step 10

Move the term $\frac{2}{5}$ outside of the limit because it is constant with respect to $x$ .

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

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.

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

Step 11.1.2

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

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

Step 11.1.3

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ approaches $\infty$ .

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

Step 11.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 11.2

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

Step 11.3

Find the derivative of the numerator and denominator.

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Step 11.3.1

Differentiate the numerator and denominator.

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

Step 11.3.2

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

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

Step 11.3.3

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 $g (x) = 5 x$ .

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Step 11.3.3.1

To apply the Chain Rule, set $u$ as $5 x$ .

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

Step 11.3.3.2

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

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

Step 11.3.3.3

Replace all occurrences of $u$ with $5 x$ .

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

Step 11.3.4

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

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

Step 11.3.5

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

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

Step 11.3.6

Multiply $5$ by $1$ .

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

Step 11.3.7

Move $5$ to the left of $e^{5 x}$ .

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

Step 12

Move the term $\frac{1}{5}$ outside of the limit because it is constant with respect to $x$ .

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

Step 13

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{e^{5 x}}$ approaches $0$ .

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

Step 14

Simplify the answer.

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Step 14.1

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

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Step 14.1.1

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

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

Step 14.1.2

Multiply $4$ by $3$ .

$\frac{12}{5 \cdot 5} \cdot \frac{2}{5} \frac{1}{5} \cdot 0$

Step 14.1.3

Multiply $5$ by $5$ .

$\frac{12}{25} \cdot \frac{2}{5} \frac{1}{5} \cdot 0$

Step 14.2

Multiply $\frac{12}{25} \cdot \frac{2}{5}$ .

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Step 14.2.1

Multiply $\frac{12}{25}$ by $\frac{2}{5}$ .

$\frac{12 \cdot 2}{25 \cdot 5} \cdot \frac{1}{5} \cdot 0$

Step 14.2.2

Multiply $12$ by $2$ .

$\frac{24}{25 \cdot 5} \cdot \frac{1}{5} \cdot 0$

Step 14.2.3

Multiply $25$ by $5$ .

$\frac{24}{125} \cdot \frac{1}{5} \cdot 0$

Step 14.3

Multiply $\frac{24}{125} \cdot \frac{1}{5}$ .

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Step 14.3.1

Multiply $\frac{24}{125}$ by $\frac{1}{5}$ .

$\frac{24}{125 \cdot 5} \cdot 0$

Step 14.3.2

Multiply $125$ by $5$ .

$\frac{24}{625} \cdot 0$

Step 14.4

Multiply $\frac{24}{625}$ by $0$ .

$0$