Limit of (a^x-1)/x

In this tutorial we shall discuss another very important formula of limit that is

\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \ln a

Let us consider the relation

\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}

Let y = {a^x} - 1, then 1 + y = {a^x}, we have

Consider the relation

1 + y = {a^x}

Taking logarithm on both sides, we have

\begin{gathered} \ln \left( {1 + y} \right) = \ln {a^x} \\ \Rightarrow \ln \left( {1 + y} \right) = x\ln a \\ \Rightarrow x = \frac{{\ln \left( {1 + y} \right)}}{{\ln a}} \\ \end{gathered}

Also \mathop {\lim }\limits_{x \to 0} y = \mathop {\lim }\limits_{x \to 0} \left( {{a^x} - 1} \right) = {a^0} - 1 = 1 - 1 = 0.

This shows that y \to 0 as x \to 0. Therefore, the given limit can be written as

\begin{gathered} \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \mathop {\lim }\limits_{y \to 0} \frac{y}{{\frac{{\ln \left( {1 + y} \right)}}{{\ln a}}}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \mathop {\lim }\limits_{y \to 0} \frac{{\ln a}}{{\frac{1}{y}\ln \left( {1 + y} \right)}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \mathop {\lim }\limits_{y \to 0} \frac{{\ln a}}{{\ln {{\left( {1 + y} \right)}^{\frac{1}{y}}}}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \frac{{\ln a}}{{\mathop {\lim }\limits_{y \to 0} \ln {{\left( {1 + y} \right)}^{\frac{1}{y}}}}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \frac{{\ln a}}{{\ln \left[ {\mathop {\lim }\limits_{y \to 0} {{\left( {1 + y} \right)}^{\frac{1}{y}}}} \right]}} \\ \end{gathered}

Using the relation \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = e, we have

 \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \frac{{\ln a}}{{\ln e}}

But \ln e = 1, we have

 \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \frac{{\ln a}}{1} = \ln a