Fundamental limit of $ (1+x)^k - 1 $ divided by x as $ x \to 0 $

The fundamental limit associated with $ (1+x)^k $ states that, for any real exponent $ k $, \[ \lim_{x \to 0} \frac{(1 + x)^k - 1}{x} = k \]

This limit captures the local behavior of the function $ (1+x)^k $ in a neighborhood of $ x = 0 $ and coincides with its derivative at $ x=0 $.

Equivalently, for sufficiently small values of $ x $, the function admits a first-order linear approximation with slope $ k $.

Proof

We aim to establish that

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

Using the identity $ e^{\ln u} = u $, we express $ (1+x)^k $ in exponential form via the natural logarithm

$$ (1+x)^k = e^{\ln \big( (1+x)^k \big)} $$

$$ (1+x)^k = e^{\ln \big( (1+x)^k \big)} = e^{ k \ln (1+x)} $$

Substituting this expression into the limit, we obtain

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

We now multiply and divide by $ k \ln (1+x) $

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

Rearranging the factors yields

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

The first two factors are well-known standard limits

$ \lim_{x \to 0} \frac{e^{ k \ln (1+x)} - 1}{ k \ln(1+x) } = 1 $, since $ \lim_{u \to 0} \frac{e^u - 1}{u} = 1 $
$ \lim_{x \to 0} \frac{ \ln(1+x) }{ x } = 1 $

Consequently, the entire expression converges to $ k $ as $ x \to 0 $

\[ \lim_{x \to 0} \underbrace{ \frac{e^{ k \ln (1+x)} - 1}{ k \ln(1+x) } }_{1} \cdot \underbrace{ \frac{ \ln(1+x) }{ x } }_{1} \cdot k = k \]

Therefore,

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

which completes the proof.