# The Number e

The number e is defined by:

$e = \lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n}.$

The number e is a transcendental number which is approximately equal to 2.718281828... The substitution $$u = {\frac{1}{n}}$$ where $$u = {\frac{1}{n}} \to 0$$ as n → ±∞, leads to another definition for e:

$e = \lim\limits_{u \to 0} {\left( {1 + u} \right)^{\frac{1}{u}}}.$

Here we meet with power expressions, in which the base and power approach to a certain number a (or to infinity). In many cases such types of limits can be calculated by taking logarithm of the function.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate the limit $\lim\limits_{n \to \infty } {\left( {1 + {\frac{1}{n}}} \right)^{n + 5}}.$

### Example 2

Find the limit $\lim\limits_{x \to \infty } {\left( {1 + {\frac{1}{x}}} \right)^{3x}}.$

### Example 3

Calculate the limit $\lim\limits_{x \to \infty } {\left( {1 + {\frac{6}{x}}} \right)^x}.$

### Example 4

Find the limit $\lim\limits_{x \to 0} \sqrt[x]{{1 + 3x}}.$

### Example 1.

Calculate the limit $\lim\limits_{n \to \infty } {\left( {1 + {\frac{1}{n}}} \right)^{n + 5}}.$

Solution.

$\lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^{n + 5}} = \lim\limits_{n \to \infty } \left[ {{{\left( {1 + \frac{1}{n}} \right)}^n}{{\left( {1 + \frac{1}{n}} \right)}^5}} \right] = \lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n} \cdot \lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^5} = e \cdot 1 = e.$

### Example 2.

Find the limit $\lim\limits_{x \to \infty } {\left( {1 + {\frac{1}{x}}} \right)^{3x}}.$

Solution.

By the product rule for limits, we obtain

$\lim\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^{3x}} = \lim\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} \cdot \lim\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} \cdot \lim\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e \cdot e \cdot e = {e^3}.$

### Example 3.

Calculate the limit $\lim\limits_{x \to \infty } {\left( {1 + {\frac{6}{x}}} \right)^x}.$

Solution.

Substituting $${\frac{6}{x}} = {\frac{1}{y}},$$ so that $$x = 6y$$ and $$y \to \infty$$ as $$x \to \infty,$$ we obtain

$\lim\limits_{x \to \infty } {\left( {1 + \frac{6}{x}} \right)^x} = \lim\limits_{y \to \infty } {\left( {1 + \frac{1}{y}} \right)^{6y}} = \lim\limits_{y \to \infty } {\left[ {{{\left( {1 + \frac{1}{y}} \right)}^y}} \right]^6} = {\left[ {\lim\limits_{y \to \infty } {{\left( {1 + \frac{1}{y}} \right)}^y}} \right]^6} = {e^6}.$

### Example 4.

Find the limit $\lim\limits_{x \to 0} \sqrt[x]{{1 + 3x}}.$

Solution.

$\lim\limits_{x \to 0} \sqrt[x]{{1 + 3x}} = \lim\limits_{x \to 0} {\left( {1 + 3x} \right)^{\frac{1}{x}}} = \lim\limits_{3x \to 0} {\left( {1 + 3x} \right)^{\frac{1}{{3x}} \cdot 3}} = \lim\limits_{3x \to 0} {\left[ {{{\left( {1 + 3x} \right)}^{\frac{1}{{3x}}}}} \right]^3} = \left[ {\lim\limits_{3x \to 0} {{\left( {1 + 3x} \right)}^{\frac{1}{{3x}}}}} \right]^3 = {e^3}.$

See more problems on Page 2.