Question bank — Infinite limits and limits at infinity — vertical - horizontal asymptotes

False. Undefined can mean a removable hole (like $\frac{x^2-9}{x-3}$ at $x=3$). A vertical asymptote needs a one-sided limit to be $\pm \infty$, not merely "no value here".

False. ${\lim }_{x\to 0}\frac{1}{x}$ does not exist (sides disagree $+\infty$ vs $-\infty$), yet $x=0$ is a vertical asymptote — one bad side is enough.

False. $+\infty$ is not a number; it is shorthand for "grows past every bound". By the finite definition the limit does not exist — we just describe how it fails.

True. The asymptote governs only long-run ($x\to \pm \infty$) behaviour. A curve like $\frac{\sin x}{x}$ crosses $y=0$ infinitely often yet still has HA $y=0$.

False. At $x=a$ the function is undefined (or explodes), so the graph never touches or crosses the line $x=a$.

False. It can have two — one from $x\to +\infty$, one from $x\to -\infty$ — as with $\frac{\sqrt{4x^2+1}}{x}\to \pm 2$.

False. Vertical asymptotes come from denominator zeros that don't cancel; a polynomial has finitely many roots, so finitely many.

False. It has no horizontal asymptote, but it may have a slant (oblique) one — see Slant (oblique) asymptotes via polynomial division.

False. Only for $p>0$. If $p<0$ it's $x^{|p|}\to \infty$; if $p=0$ it's the constant $1$.

Error: $\sqrt{x^2}=|x|$, and for $x\to -\infty$, $|x|=-x$. The correct value is $-2$; the student ignored the sign flip.

Error: numerator is also $0$ (form $\frac{0}{0}$). Cancelling gives $x+3\to 6$, so it's a removable hole at $(3,6)$, not an asymptote. See Continuity and removable discontinuities.

Error: the two sides disagree — $+\infty$ from the right, $-\infty$ from the left — so a single (two-sided) $\infty$ is wrong. Best said: the limit does not exist (though $x=0$ is still a VA). See One-sided limits.

Error: $\frac{\infty }{\infty }$ is indeterminate, not $1$. Divide top and bottom by $x^2$ to get $\frac{3}{5}$. (Or use L'Hôpital's rule for indeterminate forms.)

Error: $(x-3{)}^2\to 0^{+}$ from both sides (a square is never negative), and numerator $\to 4>0$, so both sides give $+\infty$. The limit is $+\infty$ (VA at $x=3$).

Error: the leading-coefficient rule needs equal degrees. Here top degree $3>2$, so the limit is $+\infty$ and there is no horizontal asymptote.

Error: only one side needs to be infinite, and the two sides can have different signs (e.g. $\frac{1}{x}$: $+\infty$ right, $-\infty$ left).

It makes the denominator tend to the nonzero leading constant $b_n$, so the fraction is a clean "constant over constant"; every other term becomes a constant times $1/x^k\to 0$.

A fixed nonzero top over a shrinking bottom means "how many tiny pieces fit into a fixed amount" grows without bound; the sign is fixed by the signs of top and bottom near $a$.

The denominator can approach $0$ through positive values on one side and negative on the other, flipping the sign of the whole fraction — so the two directions can blow up oppositely.

Because a limit at infinity has a precise meaning: for any $\epsilon >0$ we exhibit $N={\epsilon }^{-1/p}$ so $x>N$ forces $1/x^p<\epsilon$. Rigour is what Limits — formal epsilon-delta definition demands.

With equal degrees the leading terms both survive division by $x^n$, leaving the ratio $a_m/b_n$; with a smaller top, every surviving term has a negative power of $x$ that vanishes, leaving $0$.

After cancelling the common factor the limit is a finite number; the graph would be perfectly smooth if we filled that single point, so the height never explodes.

The square root symbol returns the nonnegative root; if $x$ is negative, $x$ itself is negative and can't be that output, so we need $|x|$ to guarantee a nonnegative result.

Yes, provided the other side ${\lim }_{x\to a^{-}}f=\pm \infty$. One infinite one-sided limit is sufficient for the line $x=a$ to be a vertical asymptote.

It is $\pi /2$; the left side ${\lim }_{x\to -\infty }\arctan x=-\pi /2$. So $\arctan$ has two horizontal asymptotes, $y=\pm \pi /2$.

$e^{-x}=1/e^x\to 0$, so HA $y=0$ on the right. (On the left, $e^{-x}\to +\infty$ — no HA and no finite value there.)

No. $\frac{0}{0}$ is indeterminate: after cancelling it may give a finite number (hole), or a leftover factor may still blow up to $\pm \infty$ (asymptote), e.g. $\frac{x-3}{(x-3{)}^2}=\frac{1}{x-3}$.

Just one line, $y=L$; two limits sharing the same value describe a single asymptote approached from both ends.

No. A non-constant polynomial runs off to $\pm \infty$ as $x\to \pm \infty$, so it never settles to a finite $L$ — no horizontal asymptote (and no vertical either).

Horizontal asymptote $y=5$ (trivially, the limit both ways is $5$), and no vertical asymptote since it never blows up.