Calculus Examples

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

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.

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Reorder $x$ and $x^{2}$ .

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

Step 1.2.2

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

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Reorder $1$ and $- 2 x^{2}$ .

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

Step 1.3.2

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

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

Step 1.3.3

Infinity divided by infinity is undefined.

Undefined

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

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.

$lim_{x \to \infty} \frac{x + x^{2}}{1 - 2 x^{2}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x + x^{2}]}{\frac{d}{d x} [1 - 2 x^{2}]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [x + x^{2}]}{\frac{d}{d x} [1 - 2 x^{2}]}$

Step 3.2

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

$lim_{x \to \infty} \frac{\frac{d}{d x} [x] + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [1 - 2 x^{2}]}$

Step 3.3

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

$lim_{x \to \infty} \frac{1 + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [1 - 2 x^{2}]}$

Step 3.4

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

$lim_{x \to \infty} \frac{1 + 2 x}{\frac{d}{d x} [1 - 2 x^{2}]}$

Step 3.5

Reorder terms.

$lim_{x \to \infty} \frac{2 x + 1}{\frac{d}{d x} [1 - 2 x^{2}]}$

Step 3.6

By the Sum Rule, the derivative of $1 - 2 x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x^{2}]$ .

$lim_{x \to \infty} \frac{2 x + 1}{\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x^{2}]}$

Step 3.7

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 3.8

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

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Step 3.8.1

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

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

Step 3.8.2

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

$lim_{x \to \infty} \frac{2 x + 1}{0 - 2 (2 x)}$

Step 3.8.3

Multiply $2$ by $- 2$ .

$lim_{x \to \infty} \frac{2 x + 1}{0 - 4 x}$

Step 3.9

Subtract $4 x$ from $0$ .

$lim_{x \to \infty} \frac{2 x + 1}{- 4 x}$

Step 4

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

$\frac{1}{- 4} lim_{x \to \infty} \frac{2 x + 1}{x}$

Step 5

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ .

$\frac{1}{- 4} lim_{x \to \infty} \frac{\frac{2 x}{x} + \frac{1}{x}}{\frac{x}{x}}$

Step 6

Evaluate the limit.

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Step 6.1

Cancel the common factor of $x$ .

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Step 6.1.1

Cancel the common factor.

$\frac{1}{- 4} lim_{x \to \infty} \frac{\frac{2 x}{x} + \frac{1}{x}}{\frac{x}{x}}$

Step 6.1.2

Divide $2$ by $1$ .

$\frac{1}{- 4} lim_{x \to \infty} \frac{2 + \frac{1}{x}}{\frac{x}{x}}$

Step 6.2

Cancel the common factor of $x$ .

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Step 6.2.1

Cancel the common factor.

$\frac{1}{- 4} lim_{x \to \infty} \frac{2 + \frac{1}{x}}{\frac{x}{x}}$

Step 6.2.2

Rewrite the expression.

$\frac{1}{- 4} lim_{x \to \infty} \frac{2 + \frac{1}{x}}{1}$

Step 6.3

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

$\frac{1}{- 4} \cdot \frac{lim_{x \to \infty} 2 + \frac{1}{x}}{lim_{x \to \infty} 1}$

Step 6.4

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

$\frac{1}{- 4} \cdot \frac{lim_{x \to \infty} 2 + lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} 1}$

Step 6.5

Evaluate the limit of $2$ which is constant as $x$ approaches $\infty$ .

$\frac{1}{- 4} \cdot \frac{2 + lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} 1}$

Step 7

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

$\frac{1}{- 4} \cdot \frac{2 + 0}{lim_{x \to \infty} 1}$

Step 8

Evaluate the limit.

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Step 8.1

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

$\frac{1}{- 4} \cdot \frac{2 + 0}{1}$

Step 8.2

Simplify the answer.

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Step 8.2.1

Move the negative in front of the fraction.

$- \frac{1}{4} \cdot \frac{2 + 0}{1}$

Step 8.2.2

Divide $2 + 0$ by $1$ .

$- \frac{1}{4} (2 + 0)$

Step 8.2.3

Add $2$ and $0$ .

$- \frac{1}{4} \cdot 2$

Step 8.2.4

Cancel the common factor of $2$ .

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Step 8.2.4.1

Move the leading negative in $- \frac{1}{4}$ into the numerator.

$\frac{- 1}{4} \cdot 2$

Step 8.2.4.2

Factor $2$ out of $4$ .

$\frac{- 1}{2 (2)} \cdot 2$

Step 8.2.4.3

Cancel the common factor.

$\frac{- 1}{2 \cdot 2} \cdot 2$

Step 8.2.4.4

Rewrite the expression.

$\frac{- 1}{2}$

Step 8.2.5

Move the negative in front of the fraction.

$- \frac{1}{2}$