Vertical asymptote behavior via one-sided limits

As $x \to 1$, the numerator $e^x \to e^1 = e$, a positive constant, while the denominator $x - 1 \to 0$. Since the numerator does not go to 0 or infinity, this is not an indeterminate form, so L'Hopital's Rule cannot and should not be used.

For $x \to 1^-$, we have $x - 1 \to 0^-$ (a small negative number), so $\frac{e}{\text{small negative}} \to -\infty$. For $x \to 1^+$, we have $x - 1 \to 0^+$, so $\frac{e}{\text{small positive}} \to +\infty$. This confirms $x=1$ is a vertical asymptote with the function diverging to $-\infty$ on the left and $+\infty$ on the right.

As $x \to -\infty$, $e^x \to 0$ while $x - 1 \to -\infty$. This gives the form $\frac{0}{-\infty}$, which is not indeterminate — a quantity approaching 0 divided by a quantity growing unboundedly in magnitude tends to 0.

Therefore $\displaystyle\lim_{x\to -\infty} \frac{e^x}{x-1} = 0$, so $y = 0$ is a horizontal asymptote as $x \to -\infty$.

As $x \to \infty$, both $e^x \to \infty$ and $x - 1 \to \infty$, so the limit has the indeterminate form $\infty/\infty$, justifying the use of L'Hopital's Rule.

Differentiating numerator and denominator separately: $\frac{d}{dx}[e^x] = e^x$ and $\frac{d}{dx}[x-1] = 1$. So the limit becomes $\displaystyle\lim_{x\to\infty} \frac{e^x}{1} = \infty$. Since the limit is infinite (not a finite number), there is no horizontal asymptote as $x \to \infty$.