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Limits

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