Horizontal asymptote as x → +∞

Divide every term in the numerator and denominator by $e^x$: $\frac{2e^x - 1}{e^x + 3} = \frac{2 - e^{-x}}{1 + 3e^{-x}}.$ This rewrites the expression so the dominant growth term cancels, similar to comparing leading terms in a rational function.

As $x \to +\infty$, $e^{-x} \to 0$, so the expression approaches $\dfrac{2 - 0}{1 + 0} = 2$. Thus the limit is $2$, and $f(x)$ has the horizontal asymptote $y = 2$ as $x \to +\infty$.

As $x \to -\infty$, $e^x \to 0$. Substitute this limiting behavior into the numerator and denominator separately: the numerator $2e^x - 1 \to 2(0) - 1 = -1$, and the denominator $e^x + 3 \to 0 + 3 = 3$.

Therefore $\lim_{x\to-\infty} \frac{2e^x-1}{e^x+3} = \frac{-1}{3}.$ This gives a horizontal asymptote of $y = -\tfrac{1}{3}$ as $x \to -\infty$, distinct from the asymptote found as $x \to +\infty$.