Calculus Examples

$lim_{x \to \infty} \frac{8 - 3 x}{9 x^{2} + 11}$

Step 1

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

Tap for more steps...

Step 1.1

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

$\frac{lim_{x \to \infty} 8 - 3 x}{lim_{x \to \infty} 9 x^{2} + 11}$

Step 1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 1.2.1

Reorder $8$ and $- 3 x$ .

$\frac{lim_{x \to \infty} - 3 x + 8}{lim_{x \to \infty} 9 x^{2} + 11}$

Step 1.2.2

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

$\frac{- \infty}{lim_{x \to \infty} 9 x^{2} + 11}$

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.

$lim_{x \to \infty} \frac{8 - 3 x}{9 x^{2} + 11} = lim_{x \to \infty} \frac{\frac{d}{d x} [8 - 3 x]}{\frac{d}{d x} [9 x^{2} + 11]}$

Step 3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [8 - 3 x]}{\frac{d}{d x} [9 x^{2} + 11]}$

Step 3.2

By the Sum Rule, the derivative of $8 - 3 x$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [- 3 x]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [8] + \frac{d}{d x} [- 3 x]}{\frac{d}{d x} [9 x^{2} + 11]}$

Step 3.3

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{0 + \frac{d}{d x} [- 3 x]}{\frac{d}{d x} [9 x^{2} + 11]}$

Step 3.4

Evaluate $\frac{d}{d x} [- 3 x]$ .

Tap for more steps...

Step 3.4.1

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

$lim_{x \to \infty} \frac{0 - 3 \frac{d}{d x} [x]}{\frac{d}{d x} [9 x^{2} + 11]}$

Step 3.4.2

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

$lim_{x \to \infty} \frac{0 - 3 \cdot 1}{\frac{d}{d x} [9 x^{2} + 11]}$

Step 3.4.3

Multiply $- 3$ by $1$ .

$lim_{x \to \infty} \frac{0 - 3}{\frac{d}{d x} [9 x^{2} + 11]}$

Step 3.5

Subtract $3$ from $0$ .

$lim_{x \to \infty} \frac{- 3}{\frac{d}{d x} [9 x^{2} + 11]}$

Step 3.6

By the Sum Rule, the derivative of $9 x^{2} + 11$ with respect to $x$ is $\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [11]$ .

$lim_{x \to \infty} \frac{- 3}{\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [11]}$

Step 3.7

Evaluate $\frac{d}{d x} [9 x^{2}]$ .

Tap for more steps...

Step 3.7.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to \infty} \frac{- 3}{9 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [11]}$

Step 3.7.2

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

$lim_{x \to \infty} \frac{- 3}{9 (2 x) + \frac{d}{d x} [11]}$

Step 3.7.3

Multiply $2$ by $9$ .

$lim_{x \to \infty} \frac{- 3}{18 x + \frac{d}{d x} [11]}$

Step 3.8

Since $11$ is constant with respect to $x$ , the derivative of $11$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{- 3}{18 x + 0}$

Step 3.9

Add $18 x$ and $0$ .

$lim_{x \to \infty} \frac{- 3}{18 x}$

Step 4

Evaluate the limit.

Tap for more steps...

Step 4.1

Cancel the common factor of $- 3$ and $18$ .

Tap for more steps...

Step 4.1.1

Factor $3$ out of $- 3$ .

$lim_{x \to \infty} \frac{3 (- 1)}{18 x}$

Step 4.1.2

Cancel the common factors.

Tap for more steps...

Step 4.1.2.1

Factor $3$ out of $18 x$ .

$lim_{x \to \infty} \frac{3 (- 1)}{3 (6 x)}$

Step 4.1.2.2

Cancel the common factor.

$lim_{x \to \infty} \frac{3 \cdot - 1}{3 (6 x)}$

Step 4.1.2.3

Rewrite the expression.

$lim_{x \to \infty} \frac{- 1}{6 x}$

Step 4.2

Move the term $\frac{1}{6}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{6} lim_{x \to \infty} \frac{- 1}{x}$

Step 5

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

$\frac{1}{6} \cdot 0$

Step 6

Multiply $\frac{1}{6}$ by $0$ .

$0$