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

1) Compute the limit of

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

x \to +∞

The limit of

\frac{1}{x}

0

⁺ when

x \to +∞

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 limit of

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

0

when

x \to +∞

For

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

, the limit takes the indeterminate form

+∞ \cdot 0

when

x

→

+∞

Using Taylor's formula:

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

for the function ln, we compute the Taylor expansion for X=

1 + \frac{1}{x}

at the point A=

1

of order N=1

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

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

\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)}^{2}}{x}

\frac{1}{x}

2) Let's restart using the Taylor expansion:

x \cdot \frac{1}{x}

1

The limit of

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

1

when

x \to +∞

So the limit of

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

\exp (1)

when

x \to +∞

3) The approximate value of:

e

equals:

\approx 2.7183

▷Answer: $e$

▷▷Numeric:

\approx 2.7183