Algebra Examples

$y = {(1 + x)}^{\frac{1}{x}}$

Step 1

Find where the expression ${(1 + x)}^{\frac{1}{x}}$ is undefined.

$x = 0$

Step 2

The vertical asymptotes occur at areas of infinite discontinuity.

No Vertical Asymptotes

Step 3

Evaluate $lim_{x \to \infty} {(1 + x)}^{\frac{1}{x}}$ to find the horizontal asymptote.

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Step 3.1

Use the properties of logarithms to simplify the limit.

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Step 3.1.1

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

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

Step 3.1.2

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

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

Step 3.2

Evaluate the limit.

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Step 3.2.1

Move the limit into the exponent.

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

Step 3.2.2

Combine $\frac{1}{x}$ and $\ln (1 + x)$ .

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

Step 3.3

Apply L'Hospital's rule.

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Step 3.3.1

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

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Step 3.3.1.1

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

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

Step 3.3.1.2

As log approaches infinity, the value goes to $\infty$ .

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

Step 3.3.1.3

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

$e^{\frac{\infty}{\infty}}$

Step 3.3.1.4

Infinity divided by infinity is undefined.

Undefined

$e^{\frac{\infty}{\infty}}$

Step 3.3.2

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.

$lim_{x \to \infty} \frac{\ln (1 + x)}{x} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (1 + x)]}{\frac{d}{d x} [x]}$

Step 3.3.3

Find the derivative of the numerator and denominator.

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Step 3.3.3.1

Differentiate the numerator and denominator.

$e^{lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (1 + x)]}{\frac{d}{d x} [x]}}$

Step 3.3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 1 + x$ .

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Step 3.3.3.2.1

To apply the Chain Rule, set $u$ as $1 + x$ .

$e^{lim_{x \to \infty} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [1 + x]}{\frac{d}{d x} [x]}}$

Step 3.3.3.2.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

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

Step 3.3.3.2.3

Replace all occurrences of $u$ with $1 + x$ .

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

Step 3.3.3.3

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

$e^{lim_{x \to \infty} \frac{\frac{1}{1 + x} (\frac{d}{d x} [1] + \frac{d}{d x} [x])}{\frac{d}{d x} [x]}}$

Step 3.3.3.4

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

$e^{lim_{x \to \infty} \frac{\frac{1}{1 + x} (0 + \frac{d}{d x} [x])}{\frac{d}{d x} [x]}}$

Step 3.3.3.5

Add $0$ and $\frac{d}{d x} [x]$ .

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

Step 3.3.3.6

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

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

Step 3.3.3.7

Multiply $\frac{1}{1 + x}$ by $1$ .

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

Step 3.3.3.8

Reorder terms.

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

Step 3.3.3.9

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

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

Step 3.3.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 3.3.5

Multiply $\frac{1}{x + 1}$ by $1$ .

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

Step 3.4

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

$e^{0}$

Step 3.5

Anything raised to $0$ is $1$ .

$1$

Step 4

List the horizontal asymptotes:

$y = 1$

Step 5

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

No Oblique Asymptotes

Step 6

This is the set of all asymptotes.

No Vertical Asymptotes

Horizontal Asymptotes: $y = 1$

No Oblique Asymptotes

Step 7