| English | 의미 |
|---|---|
| infinite limit/ˈɪnfɪnət ˈlɪmɪt/ | 무한대 극한 |
| vertical asymptote/ˈvɜːtɪkl ˈæsɪmptəʊt/ | 수직 점근선 |
When the output runs away to infinity
- Sometimes as $x\to c$ the function doesn't settle on a number — it grows without bound.
- We write $\displaystyle\lim_{x\to c}f(x)=\infty$ (or $-\infty$) to describe this runaway behavior.
- This is an infinite limit 无穷极限. Careful: "$=\infty$" is a description, not a real value — the limit technically does not exist as a finite number.
- It happens when a denominator shrinks to zero while the numerator stays away from zero.
Writing $\lim_{x\to c}f(x)=\infty$ means the limit exists as a finite number.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
It is shorthand for unbounded growth; as a finite number the limit does not exist.
Read each side separately
- Near a blow-up, the two sides often behave differently, so use one-sided infinite limits.
- For $f(x)=\dfrac1{x-2}$: as $x\to2^-$ the denominator is a tiny negative, so $f\to-\infty$.
- As $x\to2^+$ the denominator is a tiny positive, so $f\to+\infty$.
- The sign of the shrinking denominator decides which way each side rockets.
Watch it climb the asymptote
y = a / (x − b)
As $x$ nears $b$, the reciprocal shoots to $+\infty$ from one side and $-\infty$ from the other — the line $x=b$ is a vertical asymptote.
For $f(x)=\dfrac1{x-2}$, what is $\displaystyle\lim_{x\to2^-}f(x)$?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Just left of $2$, $x-2$ is a tiny negative, so $\frac{1}{\text{tiny}^-}\to-\infty$.
Spotting the vertical asymptote
- A vertical asymptote 竖直渐近线 is the line $x=c$ that the graph hugs as it shoots to $\pm\infty$.
- For a rational function, look where the denominator is zero but the numerator is not.
- $\dfrac{x+1}{x-2}$: denominator zero at $x=2$, numerator $=3\neq0$ there → vertical asymptote $x=2$.
- (If both are zero, factor first — it might be a removable hole instead.)
At what $x$-value does $f(x)=\dfrac{x+1}{x-4}$ have a vertical asymptote?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Denominator zero at $x=4$, numerator $=5\neq0$ → asymptote $x=4$.
For $\dfrac{x^2-9}{x-3}$ at $x=3$, both top and bottom are zero. This is...
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Both zero means cancel: $x+3\to6$. It is a removable hole, not an asymptote.
The line $x=c$ that a graph hugs as it shoots to $\pm\infty$ is a vertical ____.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
It marks an infinite discontinuity.
Infinite limit ⇄ asymptote
- The graph fact and the limit fact are two views of the same thing.
- "$\lim_{x\to2^+}f(x)=+\infty$" is the curve climbing the asymptote $x=2$ from the right.
- So an infinite limit at $c$ guarantees a vertical asymptote at $x=c$, and vice versa.
- Describe both: the one-sided infinite limits and the asymptote line.
A rational function has a vertical asymptote at $x=c$ when which hold?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Denominator zero, numerator nonzero → blow-up. If the numerator is also zero, it may be a removable hole.
Writing $\lim_{x\to c}f(x)=\infty$ does not mean the limit exists as a number — it is shorthand for "grows without bound." And a zero denominator alone is not automatically an asymptote: if the numerator is also zero there, cancel first — you may find a removable hole, not a blow-up.
Analyze $f(x)=\dfrac{1}{x-2}$ near $x=2$.
- $x\to2^-$: denominator $\to0^-$, so $f\to-\infty$.
- $x\to2^+$: denominator $\to0^+$, so $f\to+\infty$.
- Numerator $1\neq0$, so $x=2$ is a vertical asymptote.
- The two-sided limit does not exist (the sides go opposite ways).
An infinite limit $\lim_{x\to c}f(x)=\pm\infty$ describes a function growing without bound as $x\to c$ — a shorthand, not a finite value. Read each side separately (the sign of the shrinking denominator sets the direction). Where a rational function's denominator is zero but its numerator is not, there is a vertical asymptote $x=c$.