Find lim as x to 0 (e^x-1)/(ln (x+1))

Find $\displaystyle\lim_{x\to0}\frac{e^{x}-1}{\ln\bigl(x+1\bigr)}$.

Step 1: The form

At $x = 0$ the numerator is $e^{0}-1 = 0$ and the denominator is $\ln1 = 0$, so this is $\dfrac00$.

Step 2: Expand both to first order

\[ e^{x}-1 = x+\frac{x^{2}}{2}+\cdots, \qquad \ln\bigl(1+x\bigr) = x-\frac{x^{2}}{2}+\cdots \]

Both begin with the same term $x$; the ratio is therefore determined by what happens after it.

Step 3: Factor out $x$ and take the limit

\[ \frac{e^{x}-1}{\ln\bigl(1+x\bigr)} = \frac{x\left(1+\dfrac{x}{2}+\cdots\right)}{x\left(1-\dfrac{x}{2}+\cdots\right)} = \frac{1+\dfrac{x}{2}+\cdots}{1-\dfrac{x}{2}+\cdots} \longrightarrow 1. \]

Step 4: Verification

\[\text{By l'H\^opital: } \lim_{x\to0}\frac{e^{x}}{1/(1+x)} = \lim_{x\to0}e^{x}\bigl(1+x\bigr) = 1 \;\;\checkmark\]

\[x = 0.01:\ \frac{0.0100502}{0.0099503} = 1.010042; \qquad x = -0.01:\ \frac{-0.0099502}{-0.0100503} = 0.990042\]

\[\text{The ratio approaches } 1 \text{ from either side, at rate } 1+x+\cdots \;\;\checkmark\]