# Derivative of Logarithm at One/Proof 2

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## Theorem

Let $\ln x$ be the natural logarithm of $x$ for real $x$ where $x > 0$.

Then:

- $\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$

## Proof

\(\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x\) | \(=\) | \(\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} - \ln 1} x\) | subtract $\ln 1 = 0$ from the numerator, from Logarithm of 1 is 0 | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \intlimits {\dfrac {\d} {\d x} \ln x} {x \mathop = 1} {}\) | Definition of Derivative of Real Function at Point | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 1\) | Derivative of Natural Logarithm Function | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) |

$\blacksquare$