Question

Show that \(\mathop {{\bf{lim}}}\limits_{n \to \infty } {\left( {{\bf{1}} + \frac{x}{n}} \right)^n} = {e^x}\) for any \(x > {\bf{0}}\).

Short answer

It is proved that the value of the expression \(\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{n}} \right)^n}\) is \({e^x}\).

Step-by-step solution

Write the assumptions to solve the problem

Let \(m = \frac{n}{x}\), then \(n = mx\).

If \(n \to \infty \), then \(m \to \infty \).

Find the value of the expression \(\mathop {{\bf{lim}}}\limits_{n \to \infty } {\left( {{\bf{1}} + \frac{x}{n}} \right)^n}\)

The expression\(\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{n}} \right)^n}\) can be simplified as:

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{n}} \right)^n} &= \mathop {\lim }\limits_{m \to \infty } {\left( {1 + \frac{1}{m}} \right)^{mx}}\\ &= {\left( {\mathop {\lim }\limits_{m \to \infty } {{\left( {1 + \frac{1}{m}} \right)}^m}} \right)^x}\\ &= {\left( e \right)^x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{By}}\;{\rm{equation 6}}} \right)\\ = {e^x}\end{aligned}\)

Thus, the value of the expression \(\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{n}} \right)^n}\) is \({e^x}\).