Calculus Examples

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

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${(\frac{1 + x}{2 + x})}^{\frac{1 - \sqrt{x}}{1 - x}}$ as $e^{\ln ({(\frac{1 + x}{2 + x})}^{\frac{1 - \sqrt{x}}{1 - x}})}$ .

$lim_{x \to \infty} e^{\ln ({(\frac{1 + x}{2 + x})}^{\frac{1 - \sqrt{x}}{1 - x}})}$

Step 1.2

Expand $\ln ({(\frac{1 + x}{2 + x})}^{\frac{1 - \sqrt{x}}{1 - x}})$ by moving $\frac{1 - \sqrt{x}}{1 - x}$ outside the logarithm.

$lim_{x \to \infty} e^{\frac{1 - \sqrt{x}}{1 - x} \ln (\frac{1 + x}{2 + x})}$

Step 2

Evaluate the limit.

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Step 2.1

Move the limit into the exponent.

$e^{lim_{x \to \infty} \frac{1 - \sqrt{x}}{1 - x} \ln (\frac{1 + x}{2 + x})}$

Step 2.2

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

$e^{lim_{x \to \infty} \frac{1 - \sqrt{x}}{1 - x} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 3

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x = \sqrt{x^{2}}$ .

$e^{lim_{x \to \infty} \frac{\frac{1}{x} - \sqrt{\frac{x}{x^{2}}}}{\frac{1}{x} - \frac{x}{x}} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 4

Evaluate the limit.

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Step 4.1

Simplify each term.

$e^{lim_{x \to \infty} \frac{\frac{1}{x} - \sqrt{\frac{x}{x^{2}}}}{\frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 4.2

Cancel the common factor of $x$ and $x^{2}$ .

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Step 4.2.1

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

$e^{lim_{x \to \infty} \frac{\frac{1}{x} - \sqrt{\frac{x^{1}}{x^{2}}}}{\frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 4.2.2

Factor $x$ out of $x^{1}$ .

$e^{lim_{x \to \infty} \frac{\frac{1}{x} - \sqrt{\frac{x \cdot 1}{x^{2}}}}{\frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 4.2.3

Cancel the common factors.

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Step 4.2.3.1

Factor $x$ out of $x^{2}$ .

$e^{lim_{x \to \infty} \frac{\frac{1}{x} - \sqrt{\frac{x \cdot 1}{x \cdot x}}}{\frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 4.2.3.2

Cancel the common factor.

$e^{lim_{x \to \infty} \frac{\frac{1}{x} - \sqrt{\frac{x \cdot 1}{x \cdot x}}}{\frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 4.2.3.3

Rewrite the expression.

$e^{lim_{x \to \infty} \frac{\frac{1}{x} - \sqrt{\frac{1}{x}}}{\frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 4.3

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

$e^{\frac{lim_{x \to \infty} \frac{1}{x} - \sqrt{\frac{1}{x}}}{lim_{x \to \infty} \frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 4.4

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

$e^{\frac{lim_{x \to \infty} \frac{1}{x} - lim_{x \to \infty} \sqrt{\frac{1}{x}}}{lim_{x \to \infty} \frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 5

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

$e^{\frac{0 - lim_{x \to \infty} \sqrt{\frac{1}{x}}}{lim_{x \to \infty} \frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 6

Move the limit under the radical sign.

$e^{\frac{0 - \sqrt{lim_{x \to \infty} \frac{1}{x}}}{lim_{x \to \infty} \frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 7

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

$e^{\frac{0 - \sqrt{0}}{lim_{x \to \infty} \frac{1}{x} - 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 8

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

$e^{\frac{0 - \sqrt{0}}{lim_{x \to \infty} \frac{1}{x} - lim_{x \to \infty} 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 9

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

$e^{\frac{0 - \sqrt{0}}{0 - lim_{x \to \infty} 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 10

Evaluate the limit.

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Step 10.1

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot lim_{x \to \infty} \ln (\frac{1 + x}{2 + x})}$

Step 10.2

Move the limit inside the logarithm.

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (lim_{x \to \infty} \frac{1 + x}{2 + x})}$

Step 11

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (lim_{x \to \infty} \frac{\frac{1}{x} + \frac{x}{x}}{\frac{2}{x} + \frac{x}{x}})}$

Step 12

Evaluate the limit.

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Step 12.1

Cancel the common factor of $x$ .

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Step 12.1.1

Cancel the common factor.

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (lim_{x \to \infty} \frac{\frac{1}{x} + \frac{x}{x}}{\frac{2}{x} + \frac{x}{x}})}$

Step 12.1.2

Rewrite the expression.

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (lim_{x \to \infty} \frac{\frac{1}{x} + 1}{\frac{2}{x} + \frac{x}{x}})}$

Step 12.2

Cancel the common factor of $x$ .

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Step 12.2.1

Cancel the common factor.

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (lim_{x \to \infty} \frac{\frac{1}{x} + 1}{\frac{2}{x} + \frac{x}{x}})}$

Step 12.2.2

Rewrite the expression.

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (lim_{x \to \infty} \frac{\frac{1}{x} + 1}{\frac{2}{x} + 1})}$

Step 12.3

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (\frac{lim_{x \to \infty} \frac{1}{x} + 1}{lim_{x \to \infty} \frac{2}{x} + 1})}$

Step 12.4

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (\frac{lim_{x \to \infty} \frac{1}{x} + lim_{x \to \infty} 1}{lim_{x \to \infty} \frac{2}{x} + 1})}$

Step 13

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + lim_{x \to \infty} 1}{lim_{x \to \infty} \frac{2}{x} + 1})}$

Step 14

Evaluate the limit.

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Step 14.1

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{lim_{x \to \infty} \frac{2}{x} + 1})}$

Step 14.2

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{lim_{x \to \infty} \frac{2}{x} + lim_{x \to \infty} 1})}$

Step 14.3

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{2 lim_{x \to \infty} \frac{1}{x} + lim_{x \to \infty} 1})}$

Step 15

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{2 \cdot 0 + lim_{x \to \infty} 1})}$

Step 16

Evaluate the limit.

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Step 16.1

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

$e^{\frac{0 - \sqrt{0}}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})}$

Step 16.2

Simplify the answer.

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Step 16.2.1

Simplify the numerator.

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Step 16.2.1.1

Rewrite $0$ as $0^{2}$ .

$e^{\frac{0 - \sqrt{0^{2}}}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})}$

Step 16.2.1.2

Pull terms out from under the radical, assuming positive real numbers.

$e^{\frac{0 - 0}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})}$

Step 16.2.1.3

Multiply $- 1$ by $0$ .

$e^{\frac{0 + 0}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})}$

Step 16.2.1.4

Add $0$ and $0$ .

$e^{\frac{0}{0 - 1 \cdot 1} \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})}$

Step 16.2.2

Simplify the denominator.

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Step 16.2.2.1

Multiply $- 1$ by $1$ .

$e^{\frac{0}{0 - 1} \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})}$

Step 16.2.2.2

Subtract $1$ from $0$ .

$e^{\frac{0}{- 1} \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})}$

Step 16.2.3

Divide $0$ by $- 1$ .

$e^{0 \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})}$

Step 16.2.4

Simplify $0 \cdot \ln (\frac{0 + 1}{2 \cdot 0 + 1})$ by moving $0$ inside the logarithm.

$e^{\ln ({(\frac{0 + 1}{2 \cdot 0 + 1})}^{0})}$

Step 16.2.5

Exponentiation and log are inverse functions.

${(\frac{0 + 1}{2 \cdot 0 + 1})}^{0}$

Step 16.2.6

Add $0$ and $1$ .

${(\frac{1}{2 \cdot 0 + 1})}^{0}$

Step 16.2.7

Simplify the denominator.

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Step 16.2.7.1

Multiply $2$ by $0$ .

${(\frac{1}{0 + 1})}^{0}$

Step 16.2.7.2

Add $0$ and $1$ .

${(\frac{1}{1})}^{0}$

Step 16.2.8

Divide $1$ by $1$ .

$1^{0}$

Step 16.2.9

Anything raised to $0$ is $1$ .

$1$