Horizontal asymptote as x → +∞

Divide numerator and denominator by $7^x$: $\frac{7^x}{7^x+5} = \frac{1}{1 + 5/7^x}.$ As $x \to +\infty$, $7^x \to \infty$, so the term $5/7^x \to 0$.

Substituting this limit gives $\lim_{x\to+\infty} \frac{1}{1+5/7^x} = \frac{1}{1+0} = 1.$ Thus $f(x)$ approaches the horizontal asymptote $y=1$ as $x \to +\infty$.

As $x \to -\infty$, $7^x \to 0^+$ since an exponential with base greater than 1 decays to zero for very negative exponents.

Substituting this behavior into $f(x)$ gives $\lim_{x\to-\infty} \frac{7^x}{7^x+5} = \frac{0}{0+5} = 0.$ So $f(x)$ approaches the horizontal asymptote $y=0$ as $x \to -\infty$.

The two horizontal asymptote heights are different numbers. As x approaches positive infinity, f approaches the value one. As x approaches negative infinity, f approaches the value zero. Since one does not equal zero, the function has two distinct horizontal asymptotes, one for each direction.