Asymptotes practice questions — Honors Math

$x - 1 \to 0^+$, so $\dfrac{1}{x - 1} \to +\infty$.

$\tan x = \dfrac{\sin x}{\cos x}$ has vertical asymptotes where $\cos x = 0$, i.e. $x = \dfrac{\pi}{2} + n\pi$.

$x^2 - 4 = 0 \Rightarrow x = \pm 2$. Both vertical asymptotes.

$f(x) = x + \dfrac{1}{x}$. As $x \to \pm\infty$, $\dfrac{1}{x} \to 0$, so the slant asymptote is $y = x$.

$\lim_{x \to \infty} x^2 = \infty$, so there is no horizontal asymptote.

$\displaystyle\lim_{x \to \pm\infty} \dfrac{2x + 1}{x - 5} = 2$, so $y = 2$.

The numerator and denominator both have degree $2$, so $$\lim_{x\to\pm\infty} f(x) = \frac{3}{1} = 3.$$ The horizontal asymptote is $y=3$ (the same at $+\infty$ and $-\infty$).

Verification (sympy): `limit((3*x2+2)/(x2+5), x, oo)` and `limit(..., x, -oo)` → both $3$. ✓

$x - 3 = 0 \Rightarrow x = 3$.

The denominator vanishes at $x=3$, and the numerator ($5$) does not vanish there, so $$\lim_{x\to 3}\frac{5}{x-3} = \pm\infty.$$ The vertical asymptote is $x=3$.

Verification (sympy): `solve(x-3, x)` → $\{3\}$; the numerator at $x=3$ is $5\neq0$. ✓

Vertical: $x = \pm 1$ (two). Horizontal: $\lim f(x) = 1$ (one). Total: $3$. (No slant — the rational function has equal-degree numerator and denominator.)