Fundamental Limit of the Exponential Function at 0
When studying exponential functions, one result comes up again and again. It connects exponential growth with logarithms and plays a key role in calculus.
As \( x \) approaches 0, the ratio between the change in the function \( a^x - 1 \) and the change in the variable \( x \) approaches the natural logarithm of the base \( a \): \[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \]
Proof
Let us show that
\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \]
We assume \( a > 0 \), since both the exponential function \( a^x \) and the logarithm \( \ln a \) are defined only for positive bases. We also exclude the case \( a = 1 \), because the function would then be constant and the limit would be trivial.
To make the computation easier, we rewrite the expression \( a^x \) using the natural exponential function \( e \).
The natural logarithm is the inverse of the exponential function. Therefore, for every \( a > 0 \), we can write \( a = e^{\ln a} \).
\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \lim_{x \to 0} \frac{( e^{\ln a} )^x - 1}{x} \]
Using the laws of exponents, \( (e^u)^x = e^{ux} \), we obtain
\[ (e^{\ln a})^x = e^{x \ln a} \]
so the limit becomes
\[ \lim_{x \to 0} \frac{e^{x \ln a} - 1}{x} \]
At this point, introduce a change of variable:
\[ t = x \ln a \]
As \( x \to 0 \), we also have \( t \to 0 \), because \( \ln a \) is constant.
Moreover,
\[ x = \frac{t}{\ln a} \]
Substituting into the limit, we get
\[ \lim_{x \to 0} \frac{e^{x \ln a} - 1}{x} = \lim_{t \to 0} \frac{e^t - 1}{t / \ln a} \]
which can be rewritten as
\[ \lim_{t \to 0} \left( \frac{e^t - 1}{t} \cdot \ln a \right) \]
Since \( \ln a \) is a constant, we can factor it out of the limit:
\[ \ln a \cdot \lim_{t \to 0} \frac{e^t - 1}{t} \]
This isolates the fundamental exponential limit, a standard result in calculus:
\[ \lim_{t \to 0} \frac{e^t - 1}{t} = 1 \]
Therefore, we conclude that
\[ \ln a \cdot \underbrace{ \lim_{t \to 0} \frac{e^t - 1}{t} }_{1} = \ln a \cdot 1 = \ln a \]
Hence, the original limit is
\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \]
This completes the proof.
