A key limit involving the natural logarithm
The limit of the natural logarithm of \( (1+x) \) divided by \( x \), as \( x \to 0 \), is equal to 1: \[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1 \]
This limit captures the local behavior of the function \( \ln(1+x) \) near zero. It plays a central role in differential calculus, Taylor series expansions, and approximation methods.
Proof
Consider the limit
\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} \]
This is an indeterminate form, since as $ x \to 0 $ both the numerator and the denominator tend to zero.
\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} = \frac{0}{0} \]
To evaluate it, rewrite the expression in the equivalent form
\[ \lim_{x \to 0} \frac{1}{x} \ln(1 + x) \]
Using the logarithmic identity, we obtain
\[ \lim_{x \to 0} \ln \left( (1 + x)^{\frac{1}{x}} \right) \]
Introduce the substitution $ t = \frac{1}{x} $, so that $ x = \frac{1}{t} $. As $ x \to 0 $, we have $ t \to \infty $
\[ \lim_{t \to \infty } \ln \left( \left(1 + \frac{1}{t}\right)^{t} \right) \]
By continuity of the logarithm, this can be written as
\[ \ln \left( \lim_{t \to \infty } \left(1 + \frac{1}{t}\right)^{t} \right) \]
The limit \( \lim_{t \to \infty } \left(1 + \frac{1}{t}\right)^{t} \) is a well-known limit and is equal to $ e $.
\[ \ln \left( \lim_{t \to \infty } \left(1 + \frac{1}{t}\right)^{t} \right) = \ln ( e ) = 1 \]
Therefore, the original limit is equal to 1
\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1 \]
Alternative proof
Consider again the limit
\[ \lim_{x \to 0} \frac{\ln(1+x)}{x} \]
We proceed using a comparison argument based on inequalities.
For \( x > -1 \), the following inequality holds
\[ \frac{x}{1+x} \le \ln(1+x) \le x \]
Divide each term by \( x \). Since this is an inequality, we must distinguish between the cases where $ x $ is positive and negative.
A] Case $ x>0 $
Dividing by \( x>0 \), the inequality remains unchanged:
\[ \frac{1}{1+x} \le \frac{\ln(1+x)}{x} \le 1 \]
Let \( x \to 0^+ \):
\[ \lim_{x \to 0^+} \frac{1}{1+x} \le \lim_{x \to 0^+} \frac{\ln(1+x)}{x} \le \lim_{x \to 0^+} 1 \]
Since \( \frac{1}{1+x} \to 1 \) and \( 1 \to 1 \), we obtain
\[ 1 \le \lim_{x \to 0^+} \frac{\ln(1+x)}{x} \le 1 \]
Hence, by the squeeze theorem,
\[ \lim_{x \to 0^+} \frac{\ln(1+x)}{x} = 1 \]
B] Case $ x<0 $
Dividing by \( x<0 \), the inequality reverses direction:
\[ \frac{1}{1+x} \ge \frac{\ln(1+x)}{x} \ge 1 \]
Let \( x \to 0^- \):
\[ \lim_{x \to 0^-} \frac{1}{1+x} \ge \lim_{x \to 0^-} \frac{\ln(1+x)}{x} \ge \lim_{x \to 0^-} 1 \]
Since \( \frac{1}{1+x} \to 1 \) and \( 1 \to 1 \), we obtain
\[ 1 \ge \lim_{x \to 0^-} \frac{\ln(1+x)}{x} \ge 1 \]
Hence, by the squeeze theorem,
\[ \lim_{x \to 0^-} \frac{\ln(1+x)}{x} = 1 \]
C] Conclusion
Since the right-hand and left-hand limits both exist and are equal to 1, it follows that
\[ \lim_{x \to 0} \frac{\ln(1+x)}{x} = 1 \]
This completes the proof.
