Limit as x approaches negative infinity

Since we are taking the limit as $x \to -\infty$, we may assume $x < -1$, so $x + 1 < 0$ and therefore $|x+1| = -(x+1) = -x - 1$. This gives $f(x) = \frac{2x-7}{-x-1}.$

Divide numerator and denominator by $x$: $f(x) = \frac{2 - 7/x}{-1 - 1/x}.$ As $x \to -\infty$, the terms $7/x$ and $1/x$ both approach $0$, so the expression approaches $\dfrac{2}{-1} = -2$. Thus $\lim_{x\to-\infty} f(x) = -2$, giving the horizontal asymptote $y = -2$.

Since we are taking the limit as $x \to +\infty$, we may assume $x > -1$, so $x + 1 > 0$ and therefore $|x+1| = x+1$. This gives $f(x) = \frac{2x-7}{x+1}.$

Divide numerator and denominator by $x$: $f(x) = \frac{2 - 7/x}{1 + 1/x}.$ As $x \to +\infty$, both $7/x$ and $1/x$ approach $0$, so the expression approaches $\dfrac{2}{1} = 2$. Thus $\lim_{x\to+\infty} f(x) = 2$, giving the horizontal asymptote $y = 2$.

The two horizontal asymptotes are different lines. As x approaches negative infinity, f approaches negative two, giving the line y equals negative two. As x approaches positive infinity, f approaches two, giving the line y equals two. Because the absolute value in the denominator behaves differently for large negative versus large positive x, this function has two distinct horizontal asymptotes.