Find lim as x to 0 (frac1x- 1/(e^x-1))

Find $\displaystyle\lim_{x\to0}\left(\frac1x-\frac{1}{e^{x}-1}\right)$.

Step 1: Identify the indeterminate form

As $x\to0$ both $\dfrac1x$ and $\dfrac{1}{e^{x}-1}$ become infinite, with the same sign on each side of the origin. The difference is of the form $\infty-\infty$.

Step 2: Combine into a single fraction

\[ \frac1x-\frac{1}{e^{x}-1} = \frac{e^{x}-1-x}{x\bigl(e^{x}-1\bigr)}. \tag{1} \]

At $x = 0$ the numerator is $e^{0}-1-0 = 0$ and the denominator is $0\cdot0 = 0$: the form is now $\dfrac00$.

Step 3: Expand using Maclaurin series

\[ e^{x} = 1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots \;\Longrightarrow\; e^{x}-1-x = \frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots \tag{2} \]

\[ e^{x}-1 = x+\frac{x^{2}}{2}+\cdots \;\Longrightarrow\; x\bigl(e^{x}-1\bigr) = x^{2}+\frac{x^{3}}{2}+\cdots \tag{3} \]

Both are of order $x^{2}$, which is why the difference of the two infinities is finite.

Step 4: Form the ratio

\[ \frac{e^{x}-1-x}{x\bigl(e^{x}-1\bigr)} = \frac{x^{2}\left(\dfrac12+\dfrac{x}{6}+\cdots\right)}{x^{2}\left(1+\dfrac{x}{2}+\cdots\right)} = \frac{\dfrac12+\dfrac{x}{6}+\cdots}{1+\dfrac{x}{2}+\cdots}. \]

Step 5: Evaluate the limit

\[ \lim_{x\to0}\frac{\dfrac12+\dfrac{x}{6}+\cdots}{1+\dfrac{x}{2}+\cdots} = \frac{1/2}{1} = \frac12. \]

Step 6: Verification

\[x = 0.1:\ 0.4916681; \qquad x = -0.1:\ 0.5083319; \qquad x = 0.01:\ 0.4991667 \;\;\checkmark\]

\[\text{Dividing the series: } \frac{1}{x}-\frac{1}{e^{x}-1} = \frac12-\frac{x}{12}+\cdots \;\;\checkmark\]

\[\text{At } x = 0.1:\ 0.5-0.00833 = 0.49167 \;\;\checkmark\]