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

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