Exercises — Infinite limits and limits at infinity — vertical - horizontal asymptotes

WHAT: vertical asymptotes come from the denominator hitting $0$ while the top stays nonzero. Set $x-4=0\Rightarrow x=4$. Top is $5\ne 0$ there. Answer: vertical asymptote $x=4$. (Check the sign both ways so you know the shape: at $x\to 4^{+}$, $x-4\to 0^{+}$, so $5/0^{+}=+\infty$; at $x\to 4^{-}$, $5/0^{-}=-\infty$.)

WHY this rule: top degree $=1$, bottom degree $=1$ — equal degrees. By Engine 2 the limit is the ratio of leading coefficients. ${\lim }_{x\to \pm \infty }\frac{7x}{2x+9}=\lim \frac{7}{2+9/x}=\frac{7}{2}.$ Answer: horizontal asymptote $y=\frac{7}{2}$.

Top degree $3$, bottom degree $2$: top bigger. Numerator grows faster, so the limit is $\pm \infty$ — no leveling off. Answer: No horizontal asymptote.

WHY divide by $x^2$: highest power in the denominator is $x^2$; dividing exposes constants plus terms that die. $={\lim }_{x\to \infty }\frac{4-\frac{3}{x}+\frac{1}{x^2}}{2+\frac{1}{x}-\frac{5}{x^2}}=\frac{4-0+0}{2+0-0}=2.$ Answer: $2$ (horizontal asymptote $y=2$).

Divide by $x^2$ (the bottom's highest power): $={\lim }_{x\to \infty }\frac{\frac{6}{x}+\frac{2}{x^2}}{1+\frac{1}{x^2}}=\frac{0+0}{1+0}=0.$ Top degree $<$ bottom degree, so this had to be $0$. Answer: $0$ (horizontal asymptote $y=0$).

WHAT: top $\to 5+2=7$ (nonzero, positive). Bottom $(x-5{)}^2\to 0$, and a square is always $\ge 0$, so it approaches $0^{+}$ from both sides. Positive $÷0^{+}=+\infty$ from left and right alike. Answer: $+\infty$. Vertical asymptote $x=5$; the curve rockets up on both sides.

Top $\to 3-1=2$ (positive, nonzero). Bottom $x-3\to 0$; its sign flips across $3$.

• Right, $x\to 3^{+}$: $x-3\to 0^{+}$. Positive $÷0^{+}=+\infty$.
• Left, $x\to 3^{-}$: $x-3\to 0^{-}$. Positive $÷0^{-}=-\infty$. The two sides disagree, so the two-sided ${\lim }_{x\to 3}$ does not exist — but a single infinite one-sided limit is enough. Answer: $x=3$ is a vertical asymptote; $+\infty$ on the right, $-\infty$ on the left.

WHY be careful: $\sqrt{9x^2+4}$ contains $\sqrt{x^2}=|x|$, and for $x\to -\infty$ we have $|x|=-x$. Factor $x^2$ out of the root: $\sqrt{9x^2+4}=|x|\sqrt{9+\frac{4}{x^2}}=-x\sqrt{9+\frac{4}{x^2}}(x<0).$ Divide top and bottom by $x$: $\frac{-x\sqrt{9+4/x^2}}{2x+1}=\frac{-\sqrt{9+4/x^2}}{2+\frac{1}{x}}\to \frac{-\sqrt{9}}{2}=\frac{-3}{2}.$ Answer: $-\frac{3}{2}$ (a left-side horizontal asymptote). (On the right, $|x|=+x$ gives $+\frac{3}{2}$ — two different asymptotes.)

Factor the bottom: $x^2-x-6=(x-3)(x+2)$. Zeros at $x=3$ and $x=-2$.

• At $x=-2$: the top $x+2$ is also $0$ — a $\frac{0}{0}$ form. Cancel: $\frac{x+2}{(x-3)(x+2)}=\frac{1}{x-3}(x\ne -2).$ So $x=-2$ is a removable hole, not a wall. Its height: $\frac{1}{-2-3}=-\frac{1}{5}$, hole at $(-2,-\frac{1}{5})$.
• At $x=3$: reduced form $\frac{1}{x-3}$ has nonzero top, bottom $\to 0$ → genuine vertical asymptote. Answer: vertical asymptote only at $x=3$; removable hole at $\left(-2,-\frac{1}{5}\right)$.

Factor everything first: top $=x(2x+3)$, bottom $=(x-2)(x+2)$. No common factor → no holes. Vertical: zeros of the bottom at $x=2$ and $x=-2$, tops nonzero there → walls at $x=\pm 2$. Sign check (numerator $x(2x+3)$, roots at $0$ and $-\frac{3}{2}$):

• Near $x=2$: top $=2(7)=14>0$. Right ($x\to 2^{+}$): $(x-2)\to 0^{+},(x+2)>0$ → bottom $\to 0^{+}$ → $+\infty$. Left: bottom $\to 0^{-}$ → $-\infty$.
• Near $x=-2$: top $=(-2)(-1)=2>0$. Right ($x\to -2^{+}$): $(x-2)<0,(x+2)\to 0^{+}$ → bottom $\to 0^{-}$ → $-\infty$. Left: bottom $\to 0^{+}$ → $+\infty$. Horizontal: equal degrees $2=2$ → ratio of leading coefficients $=\frac{2}{1}=2$. So $y=2$ on both sides. Answer: vertical $x=2$ and $x=-2$; horizontal $y=2$; no holes. See the skeleton in the figure below.

WHY no HA: top degree $2>$ bottom degree $1$, so ${\lim }_{x\to \pm \infty }f=\pm \infty$ — no leveling line. Slant asymptote (see Slant (oblique) asymptotes via polynomial division): divide. $\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}.$ As $x\to \pm \infty$, the remainder $\frac{2}{x-1}\to 0$, so $f(x)$ hugs the line $y=x+1$. Answer: no horizontal asymptote; slant asymptote $y=x+1$; also a vertical asymptote at $x=1$.

As $x\to +\infty$: $e^x\to \infty$. Divide top and bottom by $e^x$: $\frac{1-e^{-x}}{1+e^{-x}}\to \frac{1-0}{1+0}=1.$ As $x\to -\infty$: $e^x\to 0$ directly, so $\frac{e^x-1}{e^x+1}\to \frac{0-1}{0+1}=-1.$ Answer: two horizontal asymptotes, $y=1$ (right) and $y=-1$ (left).

WHY the trick: this is an $\infty -\infty$ indeterminate form; multiply by the conjugate to turn it into a fraction we can divide. $\sqrt{x^2+6x}-x=\frac{(x^2+6x)-x^2}{\sqrt{x^2+6x}+x}=\frac{6x}{\sqrt{x^2+6x}+x}.$ For $x\to \infty$, $\sqrt{x^2+6x}=x\sqrt{1+6/x}$ (here $x>0$ so $|x|=x$). Divide top and bottom by $x$: $\frac{6}{\sqrt{1+6/x}+1}\to \frac{6}{\sqrt{1}+1}=\frac{6}{2}=3.$ Answer: $3$. (Equivalently the curve $y=\sqrt{x^2+6x}$ hugs the slant line $y=x+3$.)

Form check: as $x\to \infty$, $\ln x\to \infty$ and $x\to \infty$ → $\frac{\infty }{\infty }$, indeterminate. L'Hôpital (differentiate top and bottom separately): ${\lim }_{x\to \infty }\frac{\ln x}{x}={\lim }_{x\to \infty }\frac{1/x}{1}={\lim }_{x\to \infty }\frac{1}{x}=0.$ Growth reasoning: $\ln x$ grows slower than any positive power of $x$, so the denominator wins. Answer: $0$ (horizontal asymptote $y=0$ for this function).

WHAT each condition forces:

• HA $y=3$ with equal degrees → leading-coefficient ratio $=3$; simplest is degree 1 over degree 1: $f(x)=\frac{3x+b}{x-1}$ (leading ratio $3/1=3$ ✓).
• Vertical asymptote at $x=1$ → denominator $x-1$ with nonzero top there: top $=3(1)+b=3+b$, need $\ne 0$.
• Passes through origin → $f(0)=0$: $\frac{3(0)+b}{0-1}=-b=0\Rightarrow b=0$. Answer: $f(x)=\frac{3x}{x-1}$. Check: $f(0)=0$ ✓; at $x=1$ top $=3\ne 0$ → wall ✓; ${\lim }_{x\to \pm \infty }=3$ ✓.

Vertical wall ::: nonzero over something $\to 0$; sign from left/right; check the top isn't also $0$ (hole). Horizontal ceiling ::: divide by the denominator's top power; kill $1/x^k$; equal degrees → coefficient ratio. Top degree one higher ::: no HA, but a slant asymptote from polynomial division. Indeterminate $\frac{0}{0},\frac{\infty }{\infty },\infty -\infty$ ::: reshape (factor, conjugate) or L'Hôpital.