Asymptotes practice questions — A-Level Maths

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

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

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

$x-1\to 0^{+}$, so ${\displaystyle \frac{1}{x-1}}\to +\mathrm{\infty }$.

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

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

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

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

${\displaystyle \underset{x\to \pm \mathrm{\infty }}{\lim }{\displaystyle \frac{2x+1}{x-5}}=2}$, so $y=2$.

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

Verification (sympy): `solve(x-3, x)` → $\{3\}$; the numerator at $x=3$ is $5\mathrm{\ne }0$. ✓

${\lim }_{x\to \mathrm{\infty }}{e}^{-x}=0$, so $y=0$ is a horizontal asymptote at $+\mathrm{\infty }$.