Calculus

Limits and Continuity of Functions

Limits and Continuity Logo

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 \(2.718281828\ldots\) The substitution \(u = {\frac{1}{n}}\) where \(u = {\frac{1}{n}} \to 0\) as \(n \to \pm \infty,\) 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.

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