Calculus Examples

$lim_{x \to \infty} {(- 8 x)}^{2} e^{- 3 x}$

Step 1

Rewrite ${(- 8 x)}^{2} e^{- 3 x}$ as $\frac{{(- 8 x)}^{2}}{e^{3 x}}$ .

$lim_{x \to \infty} \frac{{(- 8 x)}^{2}}{e^{3 x}}$

Step 2

Apply L'Hospital's rule.

Tap for more steps...

Step 2.1

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

Tap for more steps...

Step 2.1.1

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

$\frac{lim_{x \to \infty} {(- 8 x)}^{2}}{lim_{x \to \infty} e^{3 x}}$

Step 2.1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 2.1.2.1

Rewrite the exponentiation as a product.

$\frac{lim_{x \to \infty} - 8 x (- 8 x)}{lim_{x \to \infty} e^{3 x}}$

Step 2.1.2.2

Move $x$ .

$\frac{lim_{x \to \infty} - 8 \cdot - 8 x \cdot x}{lim_{x \to \infty} e^{3 x}}$

Step 2.1.2.3

Multiply $- 8$ by $- 8$ .

$\frac{lim_{x \to \infty} 64 x \cdot x}{lim_{x \to \infty} e^{3 x}}$

Step 2.1.2.4

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

$\frac{lim_{x \to \infty} 64 (x^{1} x)}{lim_{x \to \infty} e^{3 x}}$

Step 2.1.2.5

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

$\frac{lim_{x \to \infty} 64 (x^{1} x^{1})}{lim_{x \to \infty} e^{3 x}}$

Step 2.1.2.6

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

$\frac{lim_{x \to \infty} 64 x^{1 + 1}}{lim_{x \to \infty} e^{3 x}}$

Step 2.1.2.7

Add $1$ and $1$ .

$\frac{lim_{x \to \infty} 64 x^{2}}{lim_{x \to \infty} e^{3 x}}$

Step 2.1.2.8

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

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

Step 2.1.3

Since the exponent $3 x$ approaches $\infty$ , the quantity $e^{3 x}$ approaches $\infty$ .

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

Step 2.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 2.2

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.

$lim_{x \to \infty} \frac{{(- 8 x)}^{2}}{e^{3 x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [{(- 8 x)}^{2}]}{\frac{d}{d x} [e^{3 x}]}$

Step 2.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 2.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [{(- 8 x)}^{2}]}{\frac{d}{d x} [e^{3 x}]}$

Step 2.3.2

Apply the product rule to $- 8 x$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [{(- 8)}^{2} x^{2}]}{\frac{d}{d x} [e^{3 x}]}$

Step 2.3.3

Raise $- 8$ to the power of $2$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [64 x^{2}]}{\frac{d}{d x} [e^{3 x}]}$

Step 2.3.4

Since $64$ is constant with respect to $x$ , the derivative of $64 x^{2}$ with respect to $x$ is $64 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to \infty} \frac{64 \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{3 x}]}$

Step 2.3.5

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

$lim_{x \to \infty} \frac{64 (2 x)}{\frac{d}{d x} [e^{3 x}]}$

Step 2.3.6

Multiply $2$ by $64$ .

$lim_{x \to \infty} \frac{128 x}{\frac{d}{d x} [e^{3 x}]}$

Step 2.3.7

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 $g (x) = 3 x$ .

Tap for more steps...

Step 2.3.7.1

To apply the Chain Rule, set $u$ as $3 x$ .

$lim_{x \to \infty} \frac{128 x}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x]}$

Step 2.3.7.2

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

$lim_{x \to \infty} \frac{128 x}{e^{u} \frac{d}{d x} [3 x]}$

Step 2.3.7.3

Replace all occurrences of $u$ with $3 x$ .

$lim_{x \to \infty} \frac{128 x}{e^{3 x} \frac{d}{d x} [3 x]}$

Step 2.3.8

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{128 x}{e^{3 x} (3 \frac{d}{d x} [x])}$

Step 2.3.9

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

$lim_{x \to \infty} \frac{128 x}{e^{3 x} (3 \cdot 1)}$

Step 2.3.10

Multiply $3$ by $1$ .

$lim_{x \to \infty} \frac{128 x}{e^{3 x} \cdot 3}$

Step 2.3.11

Move $3$ to the left of $e^{3 x}$ .

$lim_{x \to \infty} \frac{128 x}{3 e^{3 x}}$

Step 3

Move the term $\frac{128}{3}$ outside of the limit because it is constant with respect to $x$ .

$\frac{128}{3} lim_{x \to \infty} \frac{x}{e^{3 x}}$

Step 4

Apply L'Hospital's rule.

Tap for more steps...

Step 4.1

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

Tap for more steps...

Step 4.1.1

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

$\frac{128}{3} \cdot \frac{lim_{x \to \infty} x}{lim_{x \to \infty} e^{3 x}}$

Step 4.1.2

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

$\frac{128}{3} \cdot \frac{\infty}{lim_{x \to \infty} e^{3 x}}$

Step 4.1.3

Since the exponent $3 x$ approaches $\infty$ , the quantity $e^{3 x}$ approaches $\infty$ .

$\frac{128}{3} \cdot \frac{\infty}{\infty}$

Step 4.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{128}{3} \cdot \frac{\infty}{\infty}$

Step 4.2

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

Step 4.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 4.3.1

Differentiate the numerator and denominator.

$\frac{128}{3} lim_{x \to \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [e^{3 x}]}$

Step 4.3.2

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

$\frac{128}{3} lim_{x \to \infty} \frac{1}{\frac{d}{d x} [e^{3 x}]}$

Step 4.3.3

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 $g (x) = 3 x$ .

Tap for more steps...

Step 4.3.3.1

To apply the Chain Rule, set $u$ as $3 x$ .

$\frac{128}{3} lim_{x \to \infty} \frac{1}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x]}$

Step 4.3.3.2

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

$\frac{128}{3} lim_{x \to \infty} \frac{1}{e^{u} \frac{d}{d x} [3 x]}$

Step 4.3.3.3

Replace all occurrences of $u$ with $3 x$ .

$\frac{128}{3} lim_{x \to \infty} \frac{1}{e^{3 x} \frac{d}{d x} [3 x]}$

Step 4.3.4

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$\frac{128}{3} lim_{x \to \infty} \frac{1}{e^{3 x} \cdot 3 \frac{d}{d x} [x]}$

Step 4.3.5

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

$\frac{128}{3} lim_{x \to \infty} \frac{1}{e^{3 x} \cdot 3 \cdot 1}$

Step 4.3.6

Multiply $3$ by $1$ .

$\frac{128}{3} lim_{x \to \infty} \frac{1}{e^{3 x} \cdot 3}$

Step 4.3.7

Move $3$ to the left of $e^{3 x}$ .

$\frac{128}{3} lim_{x \to \infty} \frac{1}{3 e^{3 x}}$

Step 5

Move the term $\frac{1}{3}$ outside of the limit because it is constant with respect to $x$ .

$\frac{128}{3} \cdot \frac{1}{3} lim_{x \to \infty} \frac{1}{e^{3 x}}$

Step 6

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{e^{3 x}}$ approaches $0$ .

$\frac{128}{3} \cdot \frac{1}{3} \cdot 0$

Step 7

Simplify the answer.

Tap for more steps...

Step 7.1

Multiply $\frac{128}{3} \cdot \frac{1}{3}$ .

Tap for more steps...

Step 7.1.1

Multiply $\frac{128}{3}$ by $\frac{1}{3}$ .

$\frac{128}{3 \cdot 3} \cdot 0$

Step 7.1.2

Multiply $3$ by $3$ .

$\frac{128}{9} \cdot 0$

Step 7.2

Multiply $\frac{128}{9}$ by $0$ .

$0$