Graph (1+1/x)^x

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

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

$x = 0$

The vertical asymptotes occur at areas of infinite discontinuity.

No Vertical Asymptotes

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

Combine terms.

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Use the properties of logarithms to simplify the limit.

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

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

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

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

Move the limit into the exponent.

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

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

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

Apply L\'Hospital\'s rule.

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

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

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

Evaluate the limit of the numerator.

Move the limit inside the logarithm.

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

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

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

Evaluate the limit.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

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

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

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

Evaluate the limit.

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

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

Simplify the answer.

Divide $1 + 0$ by $1$ .

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

Add $1$ and $0$ .

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

The natural logarithm of $1$ is $0$ .

$e^{\frac{0}{lim_{x \to \infty} \frac{1}{x}}}$

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

$e^{\frac{0}{0}}$

The expression contains a division by $0$ . The expression is undefined.

Undefined

$e^{\frac{0}{0}}$

Since $\frac{0}{0}$ 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.

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

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

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

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 $g (x) = \frac{x + 1}{x}$ .

To apply the Chain Rule, set $u$ as $\frac{x + 1}{x}$ .

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

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

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

Replace all occurrences of $u$ with $\frac{x + 1}{x}$ .

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

Multiply by the reciprocal of the fraction to divide by $\frac{x + 1}{x}$ .

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

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

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

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x + 1$ and $g (x) = x$ .

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

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

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

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

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

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

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

Add $1$ and $0$ .

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

Multiply $x$ by $1$ .

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

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

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

Multiply $- 1$ by $1$ .

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

Multiply $\frac{x}{x + 1}$ by $\frac{x - (x + 1)}{x^{2}}$ .

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

Cancel the common factors.

Factor $x$ out of $(x + 1) x^{2}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

Apply the distributive property.

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

Apply the distributive property.

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

Simplify the numerator.

Subtract $x$ from $x$ .

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

Subtract $1 \cdot 1$ from $0$ .

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

Multiply $- 1$ by $1$ .

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

Combine terms.

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

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

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

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Multiply $x$ by $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

Factor $x$ out of $x \cdot x + x \cdot 1$ .

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

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

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

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Combine factors.

Multiply $- 1$ by $- 1$ .

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

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

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

Combine $\frac{1}{x (x + 1)}$ and $x^{2}$ .

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

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

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Evaluate the limit.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

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

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

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

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

$e^{\frac{1}{1 + 0}}$

Simplify the answer.

Add $1$ and $0$ .

$e^{\frac{1}{1}}$

Divide $1$ by $1$ .

$e^{1}$

Simplify.

$e$

List the horizontal asymptotes:

$y = e$

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

This is the set of all asymptotes.

No Vertical Asymptotes

Horizontal Asymptotes: $y = e$

No Oblique Asymptotes

Graph (1+1/x)^x

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