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
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} – 1}}{x} = \frac{{\ln a}}{1} = \ln a\]
Vinod
November 15 @ 8:27 am
Your proof was very helpful. I was looking for a proof of this limit everywhere to better understand ‘e’ to teach my daughter. thanks
Eduardo
January 8 @ 8:03 pm
Dear
How to calculate the same limit but x –> infinity ?
Sas
January 8 @ 6:05 pm
If you’re referring to
lim(a^x–1)/x as x->inf
then if it is +inf, then
lim(a^x–1)/x as x->+inf = lim(a^x/x – 1/x) as x->+inf
in this case a^x/x is +inf if a>1, and a^x/x is 0 if 0<a<1; 1/x is 0 either way.
It becomes analogical with -inf
SSC
April 7 @ 3:12 pm
We could also use Lim (h=>0) [ (e^h – 1)/h] = 1.
For that we write a^x as e^(x ln a).
Then the given limit becomes : [ e^(x ln a) – 1] /x.
Multiply numerator and denomenator by ln a and use the limit where h = xln a =>0.