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Limit Definition Quiz · Q5

The easy one — a linear function. The 5-step dance still works, but it collapses into almost nothing: the $h$ shows up once, cancels once, and the answer pops out. This is the cleanest test of whether the method generalizes to the trivial case.

Question 5 (1 pt) Find $f'(x)$ if $f(x) = 2x + 1$.

Choices: A. $f'(x) = 2$ B. $f'(x) = 2x$ C. $f'(x) = 2x + 2$ D. $f'(x) = 2h$

1 · The intuition

Intuition

A linear function is a straight line. The slope of a straight line is the same everywhere — that's the whole definition of "linear." So $f'(x)$ should be a constant, with no $x$ in it. That immediately rules out B and C (both have $x$ in them, which would mean the slope is changing).

It also rules out D: $2h$ depends on the "step size" $h$ that we set to zero at the end. The derivative can't depend on $h$ (otherwise it would change every time we picked a different step size, which doesn't make sense for a tangent line). So the answer has to be the pure constant $2$ — which is the slope of the line $2x + 1$. The constant $1$ in the function disappears in the derivative; only the coefficient of $x$ survives.

2 · The formula

Formula

For any function $f$, the derivative is the limit of the difference quotient:

$$f'(x) \;=\; \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

For a linear function $f(x) = mx + b$, the answer should be $m$ — the constant slope. Let's prove it with the 5-step dance.

3 · Worked solution

Worked solution

Given $f(x) = 2x + 1$. We compute $\displaystyle f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$ in five steps.

Step 1 — Write the difference quotient.

$$f'(x) \;=\; \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Step 2 — Compute $f(x+h)$ by substitution.

$$f(x+h) \;=\; 2(x+h) + 1 \;=\; 2x + 2h + 1$$

No squaring, no expansion — just distribute the $2$.

Step 3 — Subtract $f(x) = 2x + 1$.

$$f(x+h) - f(x) \;=\; (2x + 2h + 1) - (2x + 1) \;=\; 2h$$

The $2x$ cancels, the $+1$ cancels. All that's left is $2h$. Same "cancellation is the whole trick" pattern as Q1, Q2, and Q4 — the constant $1$ in the function is exactly the kind of thing that vanishes in the derivative.

Step 4 — Divide by $h$ and cancel.

$$\frac{2h}{h} \;=\; 2$$

The $h$ in the numerator matches the $h$ in the denominator — one cancellation and you're done.

Step 5 — Take the limit as $h \to 0$.

$$f'(x) \;=\; \lim_{h \to 0} 2 \;=\; 2$$

There's no $h$ left, so the limit is just the constant. (This is the moment to notice: the answer doesn't depend on $x$ — the slope is the same everywhere, which is exactly what "linear" means.)

Final answer:

$f'(x) = 2$ (Choice A ✓)

4 · Why the other choices fail

Reading the distractors

Each wrong answer maps to a specific step where the algebra got cut short, or where a wrong concept snuck in.

B. $f'(x) = 2x$ — You treated the function as if it were $f(x) = x^2$ (whose derivative is $2x$). The "$2$" in front of "$2x$" is a coefficient, not a variable. Functions and their coefficients don't follow the same rules: the $2$ stays a $2$ the whole way through, the $x$ becomes part of the $2h$ term that cancels. If you got $2x$ as an answer, the $h$ didn't cancel — you stopped at step 3 with $2h$ still hanging around but treated the $h$ as if it were $x$.
C. $f'(x) = 2x + 2$ — You applied the "power rule shortcut" for $x^2$ to the function, then added the constant's "derivative" as $+2$ instead of $0$. The constant $1$ in $f(x) = 2x + 1$ is not a variable — it shifts the function up by 1, but shifting doesn't change the slope. In the 5-step dance, the $+1$ in $f(x)$ cancels with the $+1$ in $f(x+h)$; there's no surviving constant term to add to the answer.
D. $f'(x) = 2h$ — You stopped at step 4 before dividing. $2h$ is the value of the secant slope for a particular step size $h$ — but the derivative is the limit as $h \to 0$. The whole point of step 5 is to set $h = 0$ at the end; you have to do step 5, not stop at the difference quotient. (You also probably shouldn't write $h$ in the answer at all — the derivative is a function of $x$, not of $h$.)

5 · The mental model (5 steps, every time)

Pattern

For a linear function $f(x) = mx + b$, the 5 steps collapse into something quick:

Write the difference quotient.
Compute $f(x+h) = m(x+h) + b = mx + mh + b$.
Subtract $f(x)$: $mh$ is all that's left (the $mx$ and $b$ cancel).
Divide by $h$ and cancel: $m$.
Limit: $m$ (no $h$ left).

The reason this problem is worth doing — even though the answer is obvious — is to confirm the method generalizes. If you can do the 5-step dance on a linear function and get $2$, you can trust the method on Q1's rational and Q3's square root. Linear is the cleanest test case; it's the "control" of the experiment.

6 · Check your understanding

Practice

Use the definition (the 5 steps) to find $f'(x)$ for $f(x) = -5x + 3$.

Hint: the leading coefficient is $-5$ (not $5$) and the constant is $+3$. Both signs matter — watch them carefully in steps 2 and 3.

Show answer

Step 1: $\displaystyle f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$

Step 2: $f(x+h) = -5(x+h) + 3 = -5x - 5h + 3$

Step 3: $f(x+h) - f(x) = (-5x - 5h + 3) - (-5x + 3) = -5h$

Step 4: $\dfrac{-5h}{h} = -5$

Step 5: $f'(x) = \lim_{h \to 0}(-5) = -5$

$f'(x) = -5$

Sanity check: the function $-5x + 3$ is a line with negative slope; the derivative should be negative. The constant $3$ vanishes in the derivative; only the leading coefficient $-5$ survives. ✓

7 · What's next

Three directions once you finish Q5:

Send your answer for Q5 (A, B, C, or D) and I'll check it.
Try a 6th problem of your own — pick a function and walk the 5 steps yourself. Good candidates: $f(x) = 4x - 7$ (linear), $f(x) = x^3$ (cubic — there's a fun moment in step 2 where you have to expand $(x+h)^3$), or $f(x) = 1/x^2$ at a specific point (combines Q1 and Q4 ideas).
The shortcut, later — once the 5-step dance feels automatic, the shortcut version (the power rule) will land as a "remember the dance, skip to the answer" trick. But not yet — keep doing the limit until it's muscle memory. Q5 is a great place to feel the dance at its cleanest, and then carry that confidence into harder problems.

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