Level 1 — RecognitionCalculus I — Limits & Derivatives

Calculus I — Limits & Derivatives

20 minutes30 marksprintable — key stays hidden on paper

Level 1 — Recognition Test

Time limit: 20 minutes Total marks: 30


Section A — Multiple Choice (1 mark each)

Choose the single best answer.

Q1. limx0sinxx\displaystyle\lim_{x\to 0}\frac{\sin x}{x} equals: (a) 00 (b) 11 (c) \infty (d) undefined

Q2. The derivative of xnx^n (power rule) is: (a) nxn1nx^{n-1} (b) nxn+1nx^{n+1} (c) xn+1n+1\frac{x^{n+1}}{n+1} (d) (n1)xn(n-1)x^n

Q3. limx(1+1x)x\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x} equals: (a) 11 (b) 00 (c) ee (d) \infty

Q4. The derivative of sinx\sin x is: (a) cosx\cos x (b) cosx-\cos x (c) sinx-\sin x (d) sec2x\sec^2 x

Q5. A function ff is continuous at x=ax=a if: (a) f(a)f(a) exists only (b) limxaf(x)\lim_{x\to a}f(x) exists only (c) limxaf(x)=f(a)\lim_{x\to a}f(x)=f(a) (d) ff is differentiable at aa

Q6. The derivative of exe^x is: (a) xex1xe^{x-1} (b) exe^x (c) exx\frac{e^x}{x} (d) lnx\ln x

Q7. For f(x)=1xf(x)=\dfrac{1}{x}, the line x=0x=0 is a: (a) horizontal asymptote (b) vertical asymptote (c) removable discontinuity (d) inflection point

Q8. The derivative of lnx\ln x is: (a) 1x\frac{1}{x} (b) xx (c) exe^x (d) 1xln10\frac{1}{x\ln 10}

Q9. The product rule states (uv)=(uv)'= (a) uvu'v' (b) uv+uvu'v+uv' (c) uvuvu'v-uv' (d) uvuvv2\frac{u'v-uv'}{v^2}

Q10. The difference quotient defining f(x)f'(x) is: (a) f(x+h)f(x)h\dfrac{f(x+h)-f(x)}{h} (b) f(x)f(a)x\dfrac{f(x)-f(a)}{x} (c) f(x+h)f(x)f(x+h)-f(x) (d) f(x)h\dfrac{f(x)}{h}

Q11. The Intermediate Value Theorem requires the function to be: (a) differentiable (b) continuous on a closed interval (c) increasing (d) bounded

Q12. A jump discontinuity occurs when: (a) both one-sided limits exist but are unequal (b) the limit equals the value (c) the function is undefined everywhere (d) the derivative is infinite


Section B — Matching (1 mark each; total 6)

Q13. Match each function to its derivative:

Function Derivative
(i) cosx\cos x (A) sec2x\sec^2 x
(ii) tanx\tan x (B) sinx-\sin x
(iii) axa^x (C) 11x2\frac{1}{\sqrt{1-x^2}}
(iv) arcsinx\arcsin x (D) axlnaa^x\ln a

(4 marks — one per correct pair)

Q14. Match the theorem/rule to its statement (2 marks):

Item Statement
(i) Rolle's Theorem (P) limfg=limfg\lim\frac{f}{g}=\lim\frac{f'}{g'} for 0/00/0
(ii) L'Hôpital's Rule (Q) If f(a)=f(b)f(a)=f(b), some cc has f(c)=0f'(c)=0

Section C — True/False WITH Justification (2 marks each; 1 T/F + 1 reason)

Q15. The Squeeze Theorem can be used to show limx0x2sin1x=0\displaystyle\lim_{x\to0}x^2\sin\frac1x=0. True/False — justify.

Q16. If limxaf(x)\lim_{x\to a}f(x) exists, then ff is continuous at aa. True/False — justify.

Q17. The quotient rule gives (uv)=uvuvv2\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}. True/False — justify.

Q18. At a local maximum of a differentiable function, f(x)=0f'(x)=0. True/False — justify.


Answer keyMark scheme & solutions

Section A (12 marks)

Q1. (b) 11 — the fundamental trig limit. (1)

Q2. (a) nxn1nx^{n-1} — power rule. (1)

Q3. (c) ee — definition of ee. (1)

Q4. (a) cosx\cos x — standard derivative. (1)

Q5. (c) limxaf(x)=f(a)\lim_{x\to a}f(x)=f(a) — all three conditions (limit exists, value exists, equal) summarised by this equation. (1)

Q6. (b) exe^xexe^x is its own derivative. (1)

Q7. (b) vertical asymptote — as x0x\to0, 1/x±1/x\to\pm\infty. (1)

Q8. (a) 1x\frac1x — derivative of natural log. (1)

Q9. (b) uv+uvu'v+uv' — product rule. (1)

Q10. (a) f(x+h)f(x)h\dfrac{f(x+h)-f(x)}{h} — first-principles difference quotient (limit as h0h\to0). (1)

Q11. (b) continuous on a closed interval — IVT hypothesis. (1)

Q12. (a) both one-sided limits exist but are unequal — definition of jump discontinuity. (1)

Section B (6 marks)

Q13. (i)→(B), (ii)→(A), (iii)→(D), (iv)→(C). (1 each = 4)

  • ddxcosx=sinx\frac{d}{dx}\cos x=-\sin x; ddxtanx=sec2x\frac{d}{dx}\tan x=\sec^2x; ddxax=axlna\frac{d}{dx}a^x=a^x\ln a; ddxarcsinx=11x2\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}.

Q14. (i)→(Q), (ii)→(P). (1 each = 2)

Section C (8 marks)

Q15. True. (1) Since 1sin1x1-1\le\sin\frac1x\le1, we get x2x2sin1xx2-x^2\le x^2\sin\frac1x\le x^2. Both bounds 0\to0 as x0x\to0, so by the Squeeze Theorem the middle 0\to0. (1)

Q16. False. (1) Existence of the limit is not enough — we also need f(a)f(a) defined and equal to the limit. E.g. a removable discontinuity has a limit but f(a)f(a) may differ or be undefined. (1)

Q17. True. (1) This is the correct quotient rule (numerator uvuvu'v-uv', denominator v2v^2), valid where v0v\neq0. (1)

Q18. True. (1) By Fermat's theorem, at an interior local extremum of a differentiable function the tangent is horizontal, so f(x)=0f'(x)=0 (a necessary condition). (1)

[
  {"claim":"limit sin(x)/x -> 1","code":"x=symbols('x'); result = (limit(sin(x)/x, x, 0)==1)"},
  {"claim":"limit (1+1/x)**x -> e","code":"x=symbols('x'); result = (limit((1+1/x)**x, x, oo)==E)"},
  {"claim":"derivative of tan(x) is sec(x)**2","code":"x=symbols('x'); result = simplify(diff(tan(x),x)-sec(x)**2)==0"},
  {"claim":"squeeze: limit x**2 sin(1/x) -> 0","code":"x=symbols('x'); result = (limit(x**2*sin(1/x), x, 0)==0)"}
]