Calculus I — Limits & Derivatives

Question 1

(a) (4 marks) Since $-1\le \sin(1/x)\le 1$ for all $x\ne 0$, multiply by $x^2\ge0$: $-x^2 \le x^2\sin(1/x)\le x^2. \quad(2)$ As $x\to0$, $\lim(-x^2)=0$ and $\lim(x^2)=0$. By squeeze theorem the middle term $\to 0$. (2)

(b) (5 marks) Given $\varepsilon>0$, we need $|(2x-1)-5|<\varepsilon$, i.e. $|2x-6|=2|x-3|<\varepsilon$. (2) Choose $\delta=\varepsilon/2$. (1) Then $0<|x-3|<\delta \Rightarrow |2x-6|=2|x-3|<2\delta=\varepsilon$. (2) ∎

(c) (5 marks) $f'(x)=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}.$ (1) Rationalise by multiplying by conjugate: $=\lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}=\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}. \quad(2)$ $=\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}. \quad(2)$

(d) (6 marks) Left limit: $\displaystyle\lim_{x\to0^-}\frac{\sin 2x}{x}=\lim_{x\to0^-}2\cdot\frac{\sin 2x}{2x}=2.$ (2) Right limit: $\displaystyle\lim_{x\to0^+}(bx+1)=1.$ (1) For continuity all three equal: need $2 = 1$? Impossible unless we require both sides match value $a$. Since left limit $=2$ and right limit $=1$, the two one-sided limits differ ($2\ne1$), so no value of $b$ makes $g$ continuous (the right expression at $0$ is fixed $=1$). For continuity we would need left = right = $a$; since $2\ne1$, $g$ cannot be made continuous — it has a jump discontinuity at $0$ of size $|2-1|=1$. (3) (If instead the right piece were $bx+2$ we'd get $2$; grader: award marks for correctly identifying $L^-=2$, $L^+=1$, jump. With $a=3$: the point value $3$ matches neither one-sided limit, confirming a jump/removable-type mismatch; the discontinuity remains a jump discontinuity.)

Question 2

(a) (5 marks) $v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3).$ (2) $a(t)=v'(t)=6t-12.$ (1) At rest: $t=1$ and $t=3$. (1) $v>0$ (positive direction) when $t<1$ or $t>3$; i.e. $[0,1)\cup(3,\infty)$. (1)

(b) (4 marks) Direction changes at $t=1,3$. Positions: $s(0)=2,\ s(1)=1-6+9+2=6,\ s(3)=27-54+27+2=2,\ s(4)=64-96+36+2=6.$ (2) Distances: $|6-2|+|2-6|+|6-2|=4+4+4=12$ m. (2)

(c) (5 marks) $\dfrac{dV}{dt}=4\pi r^2 \dfrac{dr}{dt}=12.$ (2) Surface area $16\pi=4\pi r^2\Rightarrow r^2=4\Rightarrow r=2.$ (1) $\dfrac{dr}{dt}=\dfrac{12}{4\pi r^2}=\dfrac{12}{4\pi\cdot4}=\dfrac{12}{16\pi}=\dfrac{3}{4\pi}\ \text{cm/s}.$ (2)

(d) (8 marks) Volume: $x^2 h = 32000 \Rightarrow h=32000/x^2.$ (1) Cost $C = 2x^2 \text{(base)} + 1\cdot(4xh)\text{(4 sides)} = 2x^2 + 4x\cdot\frac{32000}{x^2}=2x^2+\frac{128000}{x}.$ (2) $C'(x)=4x-\dfrac{128000}{x^2}=0 \Rightarrow 4x^3=128000 \Rightarrow x^3=32000\Rightarrow x=\sqrt[3]{32000}.$ (2) $x=\sqrt[3]{32000}\approx 31.75\ \text{cm}$ (exact $x=20\sqrt[3]{4}$). Then $h=32000/x^2$. With $x^3=32000$, $h=x^3/x^2=x$… check: $h=32000/x^2=x^3/x^2=x$, so $h=x\approx31.75$ cm. (1) Second derivative: $C''(x)=4+\dfrac{256000}{x^3}>0$ for $x>0$, so minimum. (2)

Question 3

(a) (4 marks) MVT: If $f$ continuous on $[a,b]$, differentiable on $(a,b)$, then $\exists c\in(a,b)$ with $f'(c)=\dfrac{f(b)-f(a)}{b-a}$. (2) $f'(x)=1/x$; $\dfrac{\ln e-\ln 1}{e-1}=\dfrac{1}{e-1}$. Set $1/c=1/(e-1)\Rightarrow c=e-1\approx1.718\in(1,e).$ (2)

(b) (6 marks) First: form $0/0$. (1) Apply L'Hôpital: $\dfrac{e^x-1}{2x}$, still $0/0$. (1) Again: $\dfrac{e^x}{2}\to \dfrac12.$ (1) Second: $x\ln x$ is $0\cdot(-\infty)$; rewrite $\dfrac{\ln x}{1/x}$ (form $-\infty/\infty$). (1) L'Hôpital: $\dfrac{1/x}{-1/x^2}=-x\to 0.$ (2)

(c)(i) (2 marks) $f(x)=x^3-x-2$, $f'(x)=3x^2-1$: $x_{n+1}=x_n-\frac{x_n^3-x_n-2}{3x_n^2-1}.$

(ii) (4 marks) $x_0=1.5$: $f=3.375-1.5-2=-0.125$; $f'=3(2.25)-1=5.75$. $x_1=1.5-(-0.125)/5.75=1.5+0.021739=1.5217.$ (2) $x_1=1.52174$: $f=3.52294-1.52174-2=0.00120$; $f'=3(2.31569)-1=5.94706.$ $x_2=1.52174-0.00120/5.94706=1.52174-0.000202=1.5215.$ (2) (True root $\approx 1.52138$.)

(iii) (2 marks)

def newton(f, df, x0, tol):
    x = x0
    while abs(f(x)) > tol:
        x = x - f(x)/df(x)
    return x

Failure condition (any one): $f'(x_n)=0$ (division by zero / flat tangent), poor initial guess causing divergence or cycling, or non-convergence when the root is a multiple root (slow) — (2).

[
  {"claim":"sqrt derivative limit equals 1/(2 sqrt x)","code":"x,h=symbols('x h',positive=True); result = simplify(limit((sqrt(x+h)-sqrt(x))/h,h,0)-1/(2*sqrt(x)))==0"},
  {"claim":"total distance travelled 0..4 is 12","code":"t=symbols('t'); s=t**3-6*t**2+9*t+2; pts=[0,1,3,4]; d=sum(abs(s.subs(t,pts[i+1])-s.subs(t,pts[i])) for i in range(3)); result = d==12"},
  {"claim":"balloon dr/dt = 3/(4 pi)","code":"r=2; drdt=12/(4*pi*r**2); result = simplify(drdt-Rational(3,4)/pi)==0"},
  {"claim":"box optimum x cubed = 32000 and second deriv positive","code":"x=symbols('x',positive=True); C=2*x**2+128000/x; sol=solve(diff(C,x),x); xc=sol[0]; result = (xc**3==32000) and (diff(C,x,2).subs(x,xc)>0)"},
  {"claim":"limits: (e^x-1-x)/x^2 -> 1/2 and x ln x -> 0","code":"x=symbols('x'); l1=limit((exp(x)-1-x)/x**2,x,0); l2=limit(x*log(x),x,0,'+'); result = (l1==Rational(1,2)) and (l2==0)"}
]