Level 3 — ProductionCalculus I — Limits & Derivatives

Calculus I — Limits & Derivatives

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level 3 Paper — Production: From-Scratch Derivations & Explain-Out-Loud

Time limit: 45 minutes Total marks: 60

Instructions: Show every step. Where a proof is requested, state assumptions and justify each line. "Explain out loud" prompts require prose reasoning, not just algebra.

Question 1 — Derivative from first principles + explain (10 marks)

(a) Using only the difference-quotient definition, derive $\dfrac{d}{dx}\left(\sqrt{x}\right)$ for $x>0$ . Show the rationalising step. (6)

(b) Explain out loud, in 2–3 sentences, why the limit $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ represents the slope of the tangent line rather than a secant line. (4)

Question 2 — Prove an important limit and use the Squeeze Theorem (12 marks)

(a) State the Squeeze Theorem precisely. (2)

(b) Prove $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$ using the geometric inequality $\sin x < x < \tan x$ for $0<x<\tfrac{\pi}{2}$ , and the Squeeze Theorem. Justify why the two-sided limit follows. (7)

Question 3 — Product & Chain rule proofs from memory (12 marks)

(a) Prove the product rule $\;(fg)' = f'g + fg'\;$ from first principles. State clearly where you add and subtract a term. (6)

(b) State the chain rule and give the sketch of its proof for the case where $g'(x)\neq 0$ near $x$ . Then use it to differentiate $y=\sin^3(2x^2+1)$ . (6)

Question 4 — L'Hôpital's rule: proof idea + application (10 marks)

(a) Using linear approximation, explain why L'Hôpital's rule works for the $\tfrac{0}{0}$ form: given $f(a)=g(a)=0$ , derive that $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f'(a)}{g'(a)}$ (assuming $g'(a)\neq 0$ and derivatives continuous). (4)

(b) Evaluate $\displaystyle\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}$ . (3)

Question 5 — Newton–Raphson from memory (code-from-memory) (8 marks)

(a) Derive the Newton–Raphson iteration formula from the linear approximation of $f$ at $x_n$ . (3)

(b) Write pseudocode (or Python) that finds a root of a given $f$ with derivative $f'$ , starting from $x_0$ , stopping when $|x_{n+1}-x_n|<10^{-8}$ or after 50 iterations. (3)

Question 6 — Optimization, real-world (8 marks)

A closed cylindrical can must hold a fixed volume $V=500\text{ cm}^3$ . Find the radius $r$ and height $h$ that minimise the total surface area. Prove your critical point is a minimum using the second derivative. (8)

Answer keyMark scheme & solutions

Question 1 (10)

(a) Let $f(x)=\sqrt x$ . $f'(x)=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}.$ Rationalise by multiplying by $\frac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}$ : (2) $=\lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt x)}=\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt x)}.$ (2) Cancel $h$ and take the limit: $\dfrac{1}{\sqrt x+\sqrt x}=\dfrac{1}{2\sqrt x}$ . (2)

(b) The quotient $\frac{f(x+h)-f(x)}{h}$ is the slope of the secant through $(x,f(x))$ and $(x+h,f(x+h))$ . As $h\to0$ the second point slides toward the first, so the secant rotates toward the limiting line touching the curve at one point — the tangent. The limit converts the average rate of change over $[x,x+h]$ into the instantaneous rate at $x$ . (4: any 2 correct ideas = 4)

Question 2 (12)

(a) If $g(x)\le f(x)\le h(x)$ near $a$ (except possibly at $a$ ) and $\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$ , then $\lim_{x\to a}f(x)=L$ . (2)

(b) For $0<x<\tfrac\pi2$ : areas give $\sin x < x < \tan x$ . Divide by $\sin x>0$ : $1<\frac{x}{\sin x}<\frac{1}{\cos x}\ \Rightarrow\ \cos x<\frac{\sin x}{x}<1.$ (3) As $x\to0^+$ , $\cos x\to1$ and $1\to1$ , so by squeeze $\lim_{x\to0^+}\frac{\sin x}{x}=1$ . (2) Since $\frac{\sin x}{x}$ is even ( $\sin(-x)/(-x)=\sin x/x$ ), the left limit equals the right, giving the two-sided limit $=1$ . (2)

(c) $\frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}=\frac{1-\cos^2x}{x^2(1+\cos x)}=\frac{\sin^2x}{x^2(1+\cos x)}=\left(\frac{\sin x}{x}\right)^2\frac{1}{1+\cos x}\to 1\cdot\frac12=\frac12.$ (3)

Question 3 (12)

(a) $(fg)'(x)=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}.$ Add and subtract $f(x+h)g(x)$ : (2) $=\lim_{h\to0}\frac{f(x+h)\big(g(x+h)-g(x)\big)+g(x)\big(f(x+h)-f(x)\big)}{h}.$ (2) $=\lim f(x+h)\cdot g'(x)+g(x)f'(x)=f(x)g'(x)+g(x)f'(x),$ using continuity of $f$ (so $f(x+h)\to f(x)$ ). (2)

(b) Chain rule: if $y=f(u)$ , $u=g(x)$ then $\frac{dy}{dx}=f'(g(x))g'(x)$ . Sketch: $\frac{f(g(x+h))-f(g(x))}{h}=\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\cdot\frac{g(x+h)-g(x)}{h}$ (valid when $g(x+h)\ne g(x)$ ); as $h\to0$ the first factor $\to f'(g(x))$ , second $\to g'(x)$ . (3) Apply: $y=\sin^3(2x^2+1)$ , $y'=3\sin^2(2x^2+1)\cdot\cos(2x^2+1)\cdot 4x=12x\sin^2(2x^2+1)\cos(2x^2+1).$ (3)

Question 4 (10)

(a) Linear approx near $a$ : $f(x)\approx f(a)+f'(a)(x-a)=f'(a)(x-a)$ since $f(a)=0$ ; similarly $g(x)\approx g'(a)(x-a)$ . Thus $\frac{f(x)}{g(x)}\approx\frac{f'(a)(x-a)}{g'(a)(x-a)}=\frac{f'(a)}{g'(a)}$ and taking $x\to a$ gives equality (with $g'(a)\ne0$ ). (4)

(b) $\frac00$ : apply L'Hôpital twice. $\lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim\frac{e^x-1}{2x}=\lim\frac{e^x}{2}=\frac12.$ (3)

(c) Rewrite $x\ln x=\frac{\ln x}{1/x}$ ( $\infty/\infty$ type). L'Hôpital: $\lim_{x\to0^+}\frac{\ln x}{1/x}=\lim\frac{1/x}{-1/x^2}=\lim(-x)=0.$ (3)

Question 5 (8)

(a) Linear approx: $f(x)\approx f(x_n)+f'(x_n)(x-x_n)$ . Set $\approx0$ and solve for the root estimate $x_{n+1}$ : $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$ (3)

(b)

def newton(f, df, x0, tol=1e-8, maxit=50):
    x = x0
    for _ in range(maxit):
        xn = x - f(x)/df(x)
        if abs(xn - x) < tol:
            return xn
        x = xn
    return x

(3: iteration formula 1, stopping condition 1, loop cap 1)

(c) $f(x)=x^2-2$ , $f'(x)=2x$ , $x_0=1.5$ : $x_1=1.5-\frac{1.5^2-2}{2(1.5)}=1.5-\frac{0.25}{3}=1.5-0.08333\ldots=1.41\overline{6}.$ (2)

Question 6 (8)

Volume: $V=\pi r^2 h=500\Rightarrow h=\frac{500}{\pi r^2}$ . (1) Surface area: $S=2\pi r^2+2\pi r h=2\pi r^2+\frac{1000}{r}$ . (2) $\frac{dS}{dr}=4\pi r-\frac{1000}{r^2}=0\Rightarrow r^3=\frac{1000}{4\pi}=\frac{250}{\pi}\Rightarrow r=\left(\frac{250}{\pi}\right)^{1/3}\approx4.301\text{ cm}.$ (2) Then $h=\frac{500}{\pi r^2}\approx\frac{500}{\pi(18.50)}\approx8.60\text{ cm}$ (so $h=2r$ ). (1) $\frac{d^2S}{dr^2}=4\pi+\frac{2000}{r^3}>0$ for all $r>0$ , so it is a minimum. (2)

[
  {"claim":"limit (1-cos x)/x^2 = 1/2","code":"x=symbols('x'); result = limit((1-cos(x))/x**2, x, 0)==Rational(1,2)"},
  {"claim":"limit (e^x-1-x)/x^2 = 1/2","code":"x=symbols('x'); result = limit((exp(x)-1-x)/x**2, x, 0)==Rational(1,2)"},
  {"claim":"limit x*ln x -> 0","code":"x=symbols('x', positive=True); result = limit(x*ln(x), x, 0, '+')==0"},
  {"claim":"Newton first iterate for x^2-2 from 1.5 is 17/12","code":"result = Rational(3,2) - (Rational(3,2)**2-2)/(2*Rational(3,2)) == Rational(17,12)"},
  {"claim":"optimal can radius cubed = 250/pi","code":"r=symbols('r', positive=True); S=2*pi*r**2+1000/r; sol=solve(diff(S,r),r); result = simplify(sol[0]**3 - 250/pi)==0"},
  {"claim":"derivative of sqrt(x) is 1/(2 sqrt x)","code":"x=symbols('x', positive=True); result = simplify(diff(sqrt(x),x) - 1/(2*sqrt(x)))==0"}
]