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

Find $\displaystyle\lim_{x\to\infty}x\left(e^{1/x}-1\right)$.

Step 1: Identify the indeterminate form

As $x\to\infty$ the exponent $\dfrac1x\to0$, so $e^{1/x}\to e^{0} = 1$ and the bracket tends to $0$, while the factor $x$ grows without bound. The product is of the form $\infty\cdot0$.

Step 2: Substitute $t = 1/x$

Let $t = \dfrac1x$, so that $x = \dfrac1t$ and $t\to0^{+}$ as $x\to+\infty$. The expression becomes

\[ x\left(e^{1/x}-1\right) = \frac{e^{t}-1}{t}, \tag{1} \]

which is the familiar $\dfrac00$ at the origin.

Step 3: Evaluate by the standard limit

From the Maclaurin expansion $e^{t} = 1+t+\dfrac{t^{2}}{2}+\cdots$, \[\frac{e^{t}-1}{t} = \frac{t+\dfrac{t^{2}}{2}+\cdots}{t} = 1+\frac{t}{2}+\cdots \longrightarrow 1 \quad\text{as } t\to0.\]

Step 4: Verification

\[x = 10:\ 1.0517092; \qquad x = 100:\ 1.0050167; \qquad x = 10^{4}:\ 1.0000500 \;\;\checkmark\]

\[\text{The series predicts } 1+\frac{1}{2x}:\ \text{at } x = 100,\ 1.0050 \;\;\checkmark\]

\[\text{Since } t\to0 \text{ from either side, the limit also holds as } x\to-\infty \;\;\checkmark\]