Single-Variable-Calculus · Unit 2 · Video 3 · Interactive Practice

The Limit Never Lands on the Point: Easy Limits, Zero over Zero, and One-Sided Limits

IKey Formulas

Formula	Name	What it tells you
$\displaystyle\lim_{x \to a} g(x) = g(a)$	Easy limit	Substitution returns a number: $\dfrac{4+3}{4^2+1} = \dfrac{7}{17}$
$\displaystyle\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$	Difference quotient	Substituting $x = x_0$ gives $\dfrac{0}{0}$ — every $f$ , every time
$\displaystyle\lim_{x \to x_0^{+}} f(x)$	Right-hand limit	Approach restricted to $x > x_0$
$\displaystyle\lim_{x \to x_0^{-}} f(x)$	Left-hand limit	Approach restricted to $x < x_0$

Key Insight: The limit at $x_0$ is computed under the standing assumption $x \neq x_0$ . That is why cancelling the factor $x - x_0$ is legal — and why the value $f(x_0)$ never enters the calculation at all.

IIVisualization 1 — Substitute, Then Read the Verdict

Substituting the limiting value either returns a number and ends the problem, or returns $\tfrac{0}{0}$ .

Limit Step 1

IIIVisualization 2 — Zero over Zero, Watched Closely

Both $\Delta f$ and $\Delta x$ shrink to zero, yet their quotient settles on $-\tfrac{1}{4}$ .

Side x 4.000

💡 Cancelling $x - 2$ is legal precisely because the limit never evaluates at $x = 2$ : the quotient and $-\tfrac{1}{2x}$ agree at every point the limit actually looks at.

IVVisualization 3 — Two Sides, and a Value That Is Separate

The two one-sided limits at $0$ are $1$ and $2$ , whatever value $f(0)$ is given.

|x| 2.200 f(0)

💡 Whether the value at a point and the limit at that point agree is a question with a name — continuity — and it is where the next video begins.

VQuiz Questions

Problem 1 · An Easy Limit

Given: $\displaystyle\lim_{x \to 2} \frac{x^2 + x}{x + 1}$ — evaluate it.

✅ Correct! Substituting gives

\frac{4 + 2}{2 + 1} = \frac{6}{3} = 2

— a meaningful number, so the limit is finished in one line.

❌ Not quite.

6

is the numerator at

x = 2

. The denominator

x + 1

becomes

3

, and the limit is the quotient of the two.

❌ Check the numerator. The numerator is

x^2 + x = 4 + 2 = 6

, not

x^2 = 4

— the

+\,x

term is substituted too.

❌ Nothing vanishes here. At

x = 2

the denominator is

x + 1 = 3

, not

0

. The indeterminate form appears only when numerator and denominator both go to

0

❌ Not quite. This is an easy limit: substitute

x = 2

into numerator and denominator separately, then divide.

Show solution

Substitute the limiting value $x = 2$ into numerator and denominator:

\lim_{x \to 2} \frac{x^2 + x}{x + 1} = \frac{2^2 + 2}{2 + 1} = \frac{6}{3} = 2

Both pieces produce ordinary numbers, and the denominator is not zero, so substitution is the answer — this is exactly what "easy limit" means.

(The quotient also simplifies, since $\dfrac{x^2 + x}{x + 1} = \dfrac{x(x+1)}{x+1} = x$ for $x \neq -1$ , which confirms the value $2$ . But no cancellation was needed to get it.)

Problem 2 · Reading the Signal $\tfrac{0}{0}$

Given: substituting $x = x_0$ into the difference quotient $\dfrac{f(x) - f(x_0)}{x - x_0}$ produces $\dfrac{0}{0}$ . What does that tell you about $f'(x_0)$ ?

✅ Correct!

\frac{0}{0}

is a signal, not a value: it says the substitution route failed. Cancel the factor

x - x_0

while

x \neq x_0

, then substitute.

❌ $\tfrac{0}{0}$ is not $0$ . Every difference quotient reads

\frac{0}{0}

on substitution — including those of

f(x) = x

f(x) = 1/x

, whose derivatives are never zero.

❌ One step too early. A derivative that exists perfectly well still reads

\frac{0}{0}

on substitution; the form appears before any computation has been done.

❌ The form says nothing about continuity. Numerator and denominator vanish together by construction, for every

f

— smooth ones included.

❌ Not quite.

\frac{0}{0}

is the starting point of a derivative computation, not its result.

Show solution

Substituting $x = x_0$ gives

\frac{f(x_0) - f(x_0)}{x_0 - x_0} = \frac{0}{0}

and this happens for every function at every point: the two differences are built to vanish together. A form that never varies can carry no information about the particular $f$ in front of you.

The way out is the standing assumption of the limit: $x \neq x_0$ throughout, so $x - x_0 \neq 0$ and it may be cancelled. Only after the cancellation does substitution produce the number $f'(x_0)$ .

Problem 3 · Cancel, Then Substitute

Given: $f(x) = x^2$ — evaluate $\displaystyle\lim_{x \to 3} \frac{f(x) - f(3)}{x - 3}$ .

✅ Correct!

\frac{x^2 - 9}{x - 3} = x + 3

for

x \neq 3

, and substituting into the cancelled form gives

6

— which is

f'(3) = 2 \cdot 3

❌ $\tfrac{0}{0}$ is not $0$ . It means the cancellation has not been done yet. Factor

x^2 - 9 = (x-3)(x+3)

and divide out the

x - 3

❌ That is $f(3)$ .

9

is the height of the graph at

x = 3

; the limit here is the slope of the tangent there.

❌ $\tfrac{0}{0}$ only says substitution failed. Since

x \neq 3

throughout the limit, the factor

x - 3

cancels and a plain number survives.

❌ Not quite. Write out

f(x) - f(3) = x^2 - 9

, factor it, cancel

x - 3

, and only then substitute.

Show solution

Step 1 — Write it out. $f(3) = 9$ , so

\frac{f(x) - f(3)}{x - 3} = \frac{x^2 - 9}{x - 3}

Step 2 — Substitution fails. At $x = 3$ this reads $\frac{9 - 9}{3 - 3} = \frac{0}{0}$ : no information.

Step 3 — Cancel while $x \neq 3$ .

\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3

Step 4 — Now substitute.

\lim_{x \to 3} (x + 3) = 6

So $f'(3) = 6$ , matching the power rule $\frac{d}{dx}x^2 = 2x$ at $x = 3$ .

Problem 4 · One Side, the Other Side, and the Value

Given: the piecewise function $g(x) = \begin{cases} 2x - 1, & x > 3 \\ 7 - x, & x \le 3 \end{cases}$ find the right-hand limit at $3$ and the value $g(3)$ .

What is $\lim_{x \to 3^{+}} g(x)$ ?

What is $g(3)$ ?

✅ Correct!

\lim_{x \to 3^{+}} g(x) = 5

while

g(3) = 4

. The value agrees with the left-hand limit and disagrees with the right-hand one — three separate computations, each done without looking at the others.

❌ Wrong branch.

4

comes from

7 - x

, which is the rule for

x \le 3

. A right-hand limit uses only points with

x > 3

❌ Close — finish the substitution.

2x - 1

x = 3

6 - 1

, not

6

❌ A jump is exactly what one-sided limits handle. Each side has its own perfectly good limit; it is the two-sided limit that fails to exist.

❌ The value need not match a limit.

5

is the right-hand limit. To get

g(3)

you read the rule that covers

x = 3

, which is

7 - x

because

3 \le 3

❌ A value is never an average of the two sides.

g(3)

is whatever the definition assigns, and here

3

falls under the branch

7 - x

❌ It is defined. The second branch carries

x \le 3

, so

x = 3

is covered:

g(3) = 7 - 3

❌ Check which branch applies. For

x \to 3^{+}

only points with

x > 3

are used, so the rule is

2x - 1

❌ Check which branch contains $x = 3$ . The condition

x \le 3

includes the point itself.

Show solution

Right-hand limit. $x \to 3^{+}$ restricts to $x > 3$ , where $g(x) = 2x - 1$ :

\lim_{x \to 3^{+}} g(x) = \lim_{x \to 3} (2x - 1) = 5

Left-hand limit. $x \to 3^{-}$ restricts to $x < 3$ , where $g(x) = 7 - x$ :

\lim_{x \to 3^{-}} g(x) = \lim_{x \to 3} (7 - x) = 4

The value. The branch $7 - x$ carries the condition $x \le 3$ , which includes $x = 3$ itself, so

g(3) = 7 - 3 = 4

Neither limit used $g(3)$ : both were computed from points beside $3$ . The two one-sided limits differ, so $\lim_{x \to 3} g(x)$ does not exist, even though $g(3)$ is perfectly well defined.

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