A Key Limit of the Exponential Function

This limit captures the behavior of the exponential function in a neighborhood of zero. \[ \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \]

For very small values of \( x \), the function \( e^x \) can be approximated by its first-order Taylor expansion:

\[ e^x = 1 + x + o(x) \]

Therefore:

\[ e^x - 1 \sim x \]

Hence, the ratio of the two quantities tends to 1.

Note. This property is fundamental because it shows that the derivative of the exponential function at zero is equal to 1. \[ (e^x)' \big|_{x=0} = 1 \] This result underpins the study of the exponential function and the definition of the number \( e \). And so on.

Proof

We evaluate the following limit:

\[ \lim_{x \to 0} \frac{e^x - 1}{x} \]

As $ x \to 0 $, the exponential function tends to 1, so $ e^x - 1 \to 0 $.

Thus, the limit takes the indeterminate form \( \frac{0}{0} \).

\[ \lim_{x \to 0} \frac{e^x - 1}{x} = \frac{0}{0} \]

To resolve this indeterminate form, introduce a temporary variable:

$$ t = e^x - 1 $$

Then the exponential function can be rewritten as

$$ e^x = t + 1 $$

Apply the natural logarithm to both sides of $ e^x = t + 1 $ and solve for $ x $:

$$ \ln( e^x ) = \ln( t + 1 ) $$

$$ x = \ln( t + 1 ) $$

Next, perform a change of variable in the limit using $ t = e^x - 1 $ and $ x = \ln( t + 1 ) $. Observe that as $ x \to 0 $, we also have $ t \to 0 $.

\[ \lim_{x \to 0} \frac{e^x - 1}{x} =   \lim_{t \to 0} \frac{t}{\ln( t + 1)} \]

Now divide both the numerator and the denominator by $ t $:

\[  \lim_{t \to 0} \frac{ \frac{ t }{t} }{ \frac{  \ln( t + 1) }{t} } \]

\[  \lim_{t \to 0} \frac{ 1 }{ \frac{  \ln( t + 1) }{t} } \]

The denominator contains a standard limit whose value is $ \lim_{t \to 0} \frac{\ln(1+t)}{t} = 1 $.

Therefore, as $ t \to 0 $, the limit evaluates to 1:

\[  \lim_{t \to 0} \frac{ 1 }{ \frac{  \ln( t + 1) }{t} } = \frac{1}{1} = 1 \]

Consequently, the original limit also tends to 1.

\[ \lim_{x \to 0} \frac{e^x - 1}{x} =    1 \]

As required.

And so on.