Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 4

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

$\frac{0 + lim_{x \to \infty} 3 x^{- 2} + lim_{x \to \infty} 1}{lim_{x \to \infty} x^{- 2} + x^{- 1} + 3}$

Step 5

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$\frac{0 + 3 lim_{x \to \infty} x^{- 2} + lim_{x \to \infty} 1}{lim_{x \to \infty} x^{- 2} + x^{- 1} + 3}$

Step 6

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{0 + 3 lim_{x \to \infty} \frac{1}{x^{2}} + lim_{x \to \infty} 1}{lim_{x \to \infty} x^{- 2} + x^{- 1} + 3}$

Step 7

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{2}}$ approaches $0$ .

$\frac{0 + 3 \cdot 0 + lim_{x \to \infty} 1}{lim_{x \to \infty} x^{- 2} + x^{- 1} + 3}$

Step 8

Evaluate the limit.

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Step 8.1

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

$\frac{0 + 3 \cdot 0 + 1}{lim_{x \to \infty} x^{- 2} + x^{- 1} + 3}$

Step 8.2

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

$\frac{0 + 3 \cdot 0 + 1}{lim_{x \to \infty} x^{- 2} + lim_{x \to \infty} x^{- 1} + lim_{x \to \infty} 3}$

Step 9

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{0 + 3 \cdot 0 + 1}{lim_{x \to \infty} \frac{1}{x^{2}} + lim_{x \to \infty} x^{- 1} + lim_{x \to \infty} 3}$

Step 10

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{2}}$ approaches $0$ .

$\frac{0 + 3 \cdot 0 + 1}{0 + lim_{x \to \infty} x^{- 1} + lim_{x \to \infty} 3}$

Step 11

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{0 + 3 \cdot 0 + 1}{0 + lim_{x \to \infty} \frac{1}{x} + lim_{x \to \infty} 3}$

Step 12

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

$\frac{0 + 3 \cdot 0 + 1}{0 + 0 + lim_{x \to \infty} 3}$

Step 13

Evaluate the limit.

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Step 13.1

Evaluate the limit of $3$ which is constant as $x$ approaches $\infty$ .

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

Step 13.2

Simplify the answer.

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Step 13.2.1

Simplify the numerator.

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Step 13.2.1.1

Multiply $3$ by $0$ .

$\frac{0 + 0 + 1}{0 + 0 + 3}$

Step 13.2.1.2

Add $0$ and $0$ .

$\frac{0 + 1}{0 + 0 + 3}$

Step 13.2.1.3

Add $0$ and $1$ .

$\frac{1}{0 + 0 + 3}$

Step 13.2.2

Simplify the denominator.

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Step 13.2.2.1

Add $0$ and $0$ .

$\frac{1}{0 + 3}$

Step 13.2.2.2

Add $0$ and $3$ .

$\frac{1}{3}$

Step 14

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{3}$

Decimal Form:

$0.\overline{3}$