Calculus I — Limits & Derivatives

1. (4 marks)

(a) Direct substitution (polynomial is continuous): $3(2)^2 - 5(2) + 1 = 12 - 10 + 1 = 3$. (2 marks)

(b) Factor: $\dfrac{(x-3)(x+3)}{x-3} = x+3 \to 6$ as $x\to 3$. (2 marks: 1 factoring, 1 answer)

2. (4 marks)

LHL: $\lim_{x\to1^-}(x+1) = 2$. (1) RHL: $\lim_{x\to1^+}(4-x) = 3$. (1) Since $2 \ne 3$, the two-sided limit does not exist (jump). (2)

3. (4 marks)

(a) Divide by $x^2$: $\dfrac{2 + 3/x - 1/x^2}{5 - 1/x + 4/x^2} \to \dfrac{2}{5}$. (2)

(b) $\dfrac{\sin 5x}{x} = 5\cdot\dfrac{\sin 5x}{5x} \to 5\cdot 1 = 5$, using $\lim_{u\to0}\frac{\sin u}{u}=1$. (2)

4. (5 marks)

$f'(x) = \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$. (1 setup)

$f(x+h) = (x+h)^2 - 3(x+h) = x^2 + 2xh + h^2 - 3x - 3h$. (1)

Difference: $f(x+h)-f(x) = 2xh + h^2 - 3h$. (1)

Quotient: $\dfrac{2xh + h^2 - 3h}{h} = 2x + h - 3$. (1)

Limit as $h\to0$: $f'(x) = 2x - 3$. (1)

5. (6 marks)

(a) $y' = 20x^4 - 6x^2 + 7$. (2)

(b) Product rule: $y' = 2x\sin x + x^2\cos x$. (2)

(c) Quotient rule: $y' = \dfrac{(1)(x+1) - x(1)}{(x+1)^2} = \dfrac{1}{(x+1)^2}$. (2)

6. (3 marks)

$y' = 4(3x^2+1)^3 \cdot (6x) = 24x(3x^2+1)^3$. (1 outer, 1 inner, 1 combine)

7. (4 marks)

Definition: $f$ is continuous at $x=a$ if (i) $f(a)$ exists, (ii) $\lim_{x\to a}f(x)$ exists, and (iii) $\lim_{x\to a}f(x) = f(a)$. (2)

For $g(x) = \frac{(x-1)(x+1)}{x-1} = x+1$ ($x\ne1$): limit at 1 is $2$ but $g(1)$ undefined → removable discontinuity. (2)

8. (4 marks)

IVT: If $f$ is continuous on $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$, then there exists $c\in(a,b)$ with $f(c)=N$. (2)

Let $f(x)=x^3-x-1$ (continuous). $f(1) = 1-1-1 = -1 < 0$, $f(2) = 8-2-1 = 5 > 0$. Since $0$ lies between $-1$ and $5$, a root exists in $(1,2)$. (2)

9. (3 marks)

Differentiate: $2x + 2y\dfrac{dy}{dx} = 0$. (2) $\dfrac{dy}{dx} = -\dfrac{x}{y}$. (1)

10. (3 marks)

(a) $y' = 3e^{3x}$. (1.5)

(b) $y' = \dfrac{2x}{x^2+1}$. (1.5)

[
  {"claim":"Q1b limit equals 6","code":"x=symbols('x'); result = limit((x**2-9)/(x-3), x, 3) == 6"},
  {"claim":"Q3a limit equals 2/5","code":"x=symbols('x'); result = limit((2*x**2+3*x-1)/(5*x**2-x+4), x, oo) == Rational(2,5)"},
  {"claim":"Q3b limit equals 5","code":"x=symbols('x'); result = limit(sin(5*x)/x, x, 0) == 5"},
  {"claim":"Q4 derivative from first principles is 2x-3","code":"x,h=symbols('x h'); f=lambda t:t**2-3*t; result = simplify(limit((f(x+h)-f(x))/h, h, 0) - (2*x-3)) == 0"},
  {"claim":"Q5c derivative is 1/(x+1)**2","code":"x=symbols('x'); result = simplify(diff(x/(x+1), x) - 1/(x+1)**2) == 0"},
  {"claim":"Q6 chain rule derivative is 24x(3x^2+1)^3","code":"x=symbols('x'); result = simplify(diff((3*x**2+1)**4, x) - 24*x*(3*x**2+1)**3) == 0"}
]