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
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.
