Both Sides Go Up

Find $\lim_{x \to 2} \frac{1}{(x-2)^2}$.

Try direct substitution: $\frac{1}{(2-2)^2} = \frac{1}{0}$: the numerator is $1$, the denominator is $0$ = Not a number: look closer

Make a table from the left: $x = 1.9$: $\frac{1}{0.01} = 100$; $x = 1.99$: $\frac{1}{0.0001} = 10000$ = Values grow without bound

Make a table from the right: $x = 2.1$: $100$; $x = 2.01$: $10000$ = Values grow without bound

Use the sign: $(x-2)^2$ is a square, so it is positive on both sides of $2$ = Both one-sided limits are $\infty$

Answer: $\lim_{x \to 2} \frac{1}{(x-2)^2} = \infty$. Both sides blow up in the same direction, so we write $\infty$ for the two-sided limit, and $x = 2$ is a vertical asymptote.

The Two Sides Disagree

Find $\lim_{x \to 2^-} \frac{1}{x-2}$, $\lim_{x \to 2^+} \frac{1}{x-2}$ and $\lim_{x \to 2} \frac{1}{x-2}$.

Left of 2: $x = 1.9$: $\frac{1}{-0.1} = -10$; $x = 1.99$: $\frac{1}{-0.01} = -100$ = $\lim_{x \to 2^-} \frac{1}{x-2} = -\infty$

Right of 2: $x = 2.1$: $\frac{1}{0.1} = 10$; $x = 2.01$: $\frac{1}{0.01} = 100$ = $\lim_{x \to 2^+} \frac{1}{x-2} = \infty$

Compare the two sides: One side goes to $-\infty$, the other to $\infty$ = No common behavior

State the two-sided limit: We cannot write $\infty$ or $-\infty$ for the two-sided limit = The two-sided limit does not exist

Answer: Left: $-\infty$, right: $\infty$. The two-sided limit $\lim_{x \to 2} \frac{1}{x-2}$ does not exist (and is not written as $\infty$). The line $x = 2$ is still a vertical asymptote.

Sign Analysis Without a Table

Find $\lim_{x \to 3^-} \frac{x+1}{x-3}$ and $\lim_{x \to 3^+} \frac{x+1}{x-3}$.

Numerator at $x = 3$: $3 + 1 = 4$ = Approaches $4$ (positive, not $0$)

Denominator at $x = 3$: $3 - 3 = 0$ = Approaches $0$: the limit is infinite

Sign of the denominator for $x < 3$: For example $x = 2.99$: $2.99 - 3 = -0.01$ = Small negative number

Left-hand limit: $\frac{\text{positive}}{\text{small negative}}$; check: $\frac{3.99}{-0.01} = -399$ = $\lim_{x \to 3^-} \frac{x+1}{x-3} = -\infty$

Right-hand limit: For $x > 3$ the denominator is a small positive number; check: $\frac{4.01}{0.01} = 401$ = $\lim_{x \to 3^+} \frac{x+1}{x-3} = \infty$

Answer: $\lim_{x \to 3^-} \frac{x+1}{x-3} = -\infty$ and $\lim_{x \to 3^+} \frac{x+1}{x-3} = \infty$. The graph has a vertical asymptote at $x = 3$, going down on the left and up on the right.

Asymptote or Hole?

Find the vertical asymptotes of $f(x) = \frac{x-2}{x^2-x-2}$ and describe the behavior near each zero of the denominator.

Factor the denominator: $x^2 - x - 2 = (x-2)(x+1)$ = Zeros at $x = 2$ and $x = -1$

Check the numerator at $x = 2$: $2 - 2 = 0$: this is the form $\frac{0}{0}$ = Not automatically an asymptote

Simplify: $\frac{x-2}{(x-2)(x+1)} = \frac{1}{x+1}$ for $x \neq 2$, so $\lim_{x \to 2} f(x) = \frac{1}{3}$ = A hole at $\left(2, \frac{1}{3}\right)$

Check the numerator at $x = -1$: $-1 - 2 = -3 \neq 0$, denominator $0$ = Vertical asymptote $x = -1$

Signs near $x = -1$: Using $\frac{1}{x+1}$: for $x > -1$ the denominator is small positive, for $x < -1$ small negative = $\lim_{x \to -1^+} f(x) = \infty$, $\lim_{x \to -1^-} f(x) = -\infty$

Answer: The only vertical asymptote is $x = -1$ (right side $\infty$, left side $-\infty$). At $x = 2$ the graph has a hole, because the limit there is the number $\frac{1}{3}$.

A Negative Numerator Over a Square

Find $\lim_{x \to -4} \frac{x-1}{(x+4)^2}$.

Numerator at $x = -4$: $-4 - 1 = -5$ = Approaches $-5$ (negative)

Denominator near $x = -4$: $(x+4)^2$ approaches $0$ and is a square, so it is positive on both sides = Small positive number on both sides

Combine the signs: $\frac{\text{negative}}{\text{small positive}}$; check: $x = -3.9$ gives $\frac{-4.9}{0.01} = -490$ = Both one-sided limits are $-\infty$

Answer: $\lim_{x \to -4} \frac{x-1}{(x+4)^2} = -\infty$. Both sides go down, so the two-sided limit is written as $-\infty$, and $x = -4$ is a vertical asymptote.

Mistake: Saying "the limit exists and equals $\infty$"

Why: $\infty$ is not a real number. Writing $= \infty$ only describes the way the function grows; the limit does not exist as a number.

Correct: Say "$f(x)$ grows without bound as $x \to a$" or "the limit is $\infty$", and remember that a limit equal to $\infty$ is not a limit that exists.

Mistake: Writing $\lim_{x \to 2} \frac{1}{x-2} = \infty$

Why: From the left the values go to $-\infty$, from the right to $\infty$. A two-sided limit can only be written as $\infty$ when both sides go to $\infty$.

Correct: Give both one-sided limits ($-\infty$ and $\infty$) and say that the two-sided limit does not exist.

Mistake: Calling every zero of the denominator a vertical asymptote

Why: If the numerator is also $0$ there, the expression has the form $\frac{0}{0}$ and may simplify. Then the limit can be a number and the graph only has a hole.

Correct: Check the numerator first: denominator $0$ and numerator not $0$ means a vertical asymptote; both $0$ means simplify and look again.

Mistake: Choosing the sign from the denominator only

Why: A negative numerator flips the result: $\frac{-5}{\text{small positive}}$ goes to $-\infty$, not $\infty$.

Correct: Find the sign of the numerator and the sign of the denominator on the side you are looking at, then combine them like in division.

Electric Force at Short Distances

The force between two charged particles is inversely proportional to the square of their distance, so it grows without bound as the distance shrinks to zero.

With $F(r) = \frac{100}{r^2}$ (in newtons, $r$ in meters), halving the distance makes the force four times larger, and $\lim_{r \to 0^+} F(r) = \infty$.

Removing the Last Impurities

Water treatment gets harder and harder as you approach complete purity. Models of the effort often have a vertical asymptote at 100%.

If filtering $p$ percent of the impurities takes $T(p) = \frac{50p}{100-p}$ minutes, then $T(50) = 50$ minutes, but $T(90) = 450$ minutes.

Images Formed by a Lens

For a lens with focal length $f$, the image distance is $d_i = \frac{f \, d_o}{d_o - f}$. As the object moves toward the focal point, the image moves infinitely far away.

For $f = 10$ cm: an object at $d_o = 20$ cm gives an image at $d_i = \frac{200}{10} = 20$ cm, and $\lim_{d_o \to 10^+} d_i = \infty$.

Q: If the limit is $\infty$, does the limit exist?

A: No. A limit exists only when it is a real number. Writing $\lim_{x \to a} f(x) = \infty$ is a more precise way of saying "the limit does not exist because the values grow without bound in the positive direction".

Q: How is an infinite limit different from a limit at infinity?

A: In an infinite limit, $x$ approaches a number and $f(x)$ blows up: $\lim_{x \to 2} \frac{1}{(x-2)^2} = \infty$. In a limit at infinity, $x$ itself grows without bound: $\lim_{x \to \infty} \frac{1}{x} = 0$. The first gives vertical asymptotes, the second horizontal ones.

Q: Can a graph cross a vertical asymptote?

A: No. At a vertical asymptote $x = a$ the function values blow up near $a$, and $f(a)$ itself is usually not defined. (Horizontal asymptotes are different: a graph can cross those.)

Q: Why does an even power in the denominator give the same sign on both sides?

A: A square (or any even power) is never negative. Near $a$, $(x-a)^2$ is a small positive number on both sides, so the sign of the quotient depends only on the numerator.