# Limit of (a^x-1)/x

In this tutorial we shall discuss another very important formula of limits, $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} – 1}}{x} = \ln a$

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
$1 + y = {a^x}$

Using the 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
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} – 1}}{x} = \frac{{\ln a}}{1} = \ln a$