limit x->infinity (1+1/x)^x — step-by-step solution

1) Let's evaluate the limit of

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

when

x \to +∞

ⓘ A number divided by a quantity that grows without bound tends to 0

The limit of

\frac{1}{x}

0

⁺ when

x \to +∞

ⓘ The limit of a sum of finite limits is the sum of the limits

The limit of

1 + \frac{1}{x}

1

when

x \to +∞

For

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

, the limit takes the indeterminate form

1^{+∞}

when

x

→

+∞

1^{±∞} case: rewrite

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

\exp (x \cdot \ln (1 + \frac{1}{x}))

, then compute the limit of

x \cdot \ln (1 + \frac{1}{x})

ⓘ The function ln is continuous, its limit is its value at the limit of its argument

The limit of

\ln (1 + \frac{1}{x})

\ln (1)

when

x \to +∞

Let's evaluate:

\ln (1)

ⓘ ln(1) = 0, since exp(0) = 1:

\ln (1)

0

0

For

x \cdot \ln (1 + \frac{1}{x})

, the limit takes the indeterminate form

+∞ \cdot 0

when

x

→

+∞

Apply the Taylor development

\ln (A) + \sum_{k = 1}^{N} {(-1)}^{k + 1} \cdot \frac{{(X - A)}^{k}}{k A^{k}}

of ln with X=

1 + \frac{1}{x}

around A=

1

(order N=1)

\ln (1) + \sum_{k = 1}^{1} {(-1)}^{k + 1} \cdot \frac{{(1 + \frac{1}{x} - 1)}^{k}}{k \cdot 1^{k}}

ⓘ ln(1) = 0, since exp(0) = 1:

\ln (1)

0

■Parentheses are computed first:

1 - 1

0

0 + \sum_{k = 1}^{1} {(-1)}^{k + 1} \cdot \frac{{(0 + \frac{1}{x})}^{k}}{k}

ⓘ Adding 0 changes nothing:

0 + \frac{1}{x}

\frac{1}{x}

0 + \sum_{k = 1}^{1} {(-1)}^{k + 1} \cdot \frac{{(\frac{1}{x})}^{k}}{k}

\sum_{k = 1}^{1} {(-1)}^{k + 1} \cdot \frac{{(\frac{1}{x})}^{k}}{k}

■Multiplying a fraction by a term multiplies its numerator:

{(-1)}^{k + 1} \cdot \frac{{(\frac{1}{x})}^{k}}{k}

\frac{{(-1)}^{k + 1} \cdot {(\frac{1}{x})}^{k}}{k}

\sum_{k = 1}^{1} \frac{{(-1)}^{k + 1} \cdot {(\frac{1}{x})}^{k}}{k}

ⅰ

\sum_{k = 1}^{1} \frac{{(-1)}^{k + 1} \cdot {(\frac{1}{x})}^{k}}{k}

\frac{{(-1)}^{1 + 1} \cdot {(\frac{1}{x})}^{1}}{1}

\frac{{(-1)}^{1 + 1} \cdot {(\frac{1}{x})}^{1}}{1}

ⓘ Multiplying a fraction by a term multiplies its numerator:

{(-1)}^{1 + 1} \cdot \frac{1}{x}

\frac{{(-1)}^{1 + 1} \cdot 1}{x}

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

■The exponent is computed before the power:

1 + 1

2

\frac{{(-1)}^{2}}{x}

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-1)}^{2}

(-1) \cdot (-1)

1

\frac{1}{x}

2) Let's restart using the Taylor expansion:

x \cdot \frac{1}{x}

ⓘ Multiplying a fraction by a term multiplies its numerator:

x \cdot \frac{1}{x}

\frac{x \cdot 1}{x}

\frac{x}{x}

■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor

x

\frac{x}{x}

1

1

The limit of

x \cdot \ln (1 + \frac{1}{x})

1

when

x \to +∞

So the limit of

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

e

when

x \to +∞

▶Limit:

e

3) The numeric value of:

e

is:

▷

\approx 2.7183