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Dr. Shreyank Educare
Curriculum-aligned learning support — not an official school, university, or ministry-approved course, and not a replacement for official credit.
Adaptive practice for this skillA separate, individually graded pool that adapts to your answers: 24 validated questions, plus generated practice on the same skill.

Picture it first

Interactive · drag the slider

Watch the limit happen

Drag x toward 3. The point climbs the line toward the open hole at (3, 6) — and never has to reach it.

234y = 6 (the limit)f(2.50) = 5.50
x = 23 · the limit pointx = 4
x
2.500
f(x)
5.500
gap to 6
0.500

You are approaching from the left. Keep sliding toward 3 and watch f(x) close in on 6.

x → 3⁻f(x)x → 3⁺f(x)
2.95.93.16.1
2.995.993.016.01
2.9995.9993.0016.001
↓ both columns close in on 6 ↓

Grade 12 · Practice problems

Limits

30 written problems across three levels — 10 foundational · 12 intermediate · 8 advanced. Work through them on paper, then reveal every answer to check. These are not graded and are counted separately from adaptive practice.

    Foundational
  1. 1. FoundationalEvaluate lim⁡x→−1(2x3−x+4)\lim_{x\to-1}(2x^3-x+4).

  2. 2. FoundationalEvaluate lim⁡x→4(x2−5x+2)\lim_{x\to4}(x^2-5x+2).

  3. 3. FoundationalEvaluate lim⁡x→0x2−3xx\lim_{x\to0}\dfrac{x^2-3x}{x}.

  4. 4. FoundationalEvaluate lim⁡x→3x2−9x−3\lim_{x\to3}\dfrac{x^2-9}{x-3}.

  5. 5. FoundationalEvaluate lim⁡x→5x−5x2−25\lim_{x\to5}\dfrac{x-5}{x^2-25}.

  6. 6. FoundationalState the three conditions for continuity at x=ax=a.

  7. 7. FoundationalIf the left-hand limit is 22 and the right-hand limit is 55, determine the two-sided limit.

  8. 8. FoundationalClassify a discontinuity whose finite one-sided limits are unequal.

  9. 9. FoundationalFor f(x)=x+1f(x)=x+1 when x<2x<2, evaluate lim⁡x→2−f(x)\lim_{x\to2^-}f(x).

  10. 10. FoundationalA graph has a hole at (4,7)(4,7) and a filled point at (4,1)(4,1), approaching the hole from both sides. Find lim⁡x→4f(x)\lim_{x\to4}f(x) and f(4)f(4).

  11. Intermediate
  12. 11. IntermediateEvaluate lim⁡x→0x+9−3x\lim_{x\to0}\dfrac{\sqrt{x+9}-3}{x}.

  13. 12. IntermediateEvaluate lim⁡x→2x2−4x2−x−2\lim_{x\to2}\dfrac{x^2-4}{x^2-x-2}.

  14. 13. IntermediateEvaluate lim⁡x→1x3−1x−1\lim_{x\to1}\dfrac{x^3-1}{x-1}.

  15. 14. IntermediateEvaluate lim⁡x→−2x2−4x+2\lim_{x\to-2}\dfrac{x^2-4}{x+2}.

  16. 15. IntermediateEvaluate lim⁡x→0(x+3)2−9x\lim_{x\to0}\dfrac{(x+3)^2-9}{x}.

  17. 16. IntermediateEvaluate lim⁡x→4x−2x−4\lim_{x\to4}\dfrac{\sqrt{x}-2}{x-4}.

  18. 17. IntermediateEvaluate lim⁡x→01x(1x+2−12)\lim_{x\to0}\dfrac{1}{x}\left(\dfrac{1}{x+2}-\dfrac{1}{2}\right).

  19. 18. IntermediateEvaluate lim⁡x→2x−2∣x−2∣\lim_{x\to2}\dfrac{x-2}{|x-2|}, or state that it does not exist.

  20. 19. IntermediateEvaluate lim⁡x→0x2cos⁡(3/x)\lim_{x\to0}x^2\cos(3/x) using the squeeze theorem.

  21. 20. IntermediateWhat value of f(2)f(2) makes f(x)=x2−4x−2f(x)=\dfrac{x^2-4}{x-2} (for x≠2x\ne2) continuous at x=2x=2?

  22. 21. IntermediateEvaluate lim⁡x→3+1x−3\lim_{x\to3^+}\dfrac{1}{x-3} and classify the discontinuity at x=3x=3.

  23. 22. IntermediateFor f(x)=x+2f(x)=x+2 when x<1x<1 and f(x)=5−xf(x)=5-x when x≥1x\ge1, evaluate lim⁡x→1f(x)\lim_{x\to1}f(x).

  24. Advanced
  25. 23. AdvancedEvaluate lim⁡x→01+x−1−xx\lim_{x\to0}\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}.

  26. 24. AdvancedFind aa so that f(x)=x2+1f(x)=x^2+1 for x<2x<2 and f(x)=ax−3f(x)=ax-3 for x≥2x\ge2 is continuous at x=2x=2.

  27. 25. AdvancedUsing one-sided limits, justify why lim⁡x→2x+3x−2\lim_{x\to2}\dfrac{x+3}{x-2} does not exist.

  28. 26. AdvancedEvaluate lim⁡x→01x+1−1x\lim_{x\to0}\dfrac{\frac{1}{x+1}-1}{x}.

  29. 27. AdvancedUse the squeeze theorem to evaluate lim⁡x→0xsin⁡(1/x)\lim_{x\to0}x\sin(1/x), and state the two bounding functions.

  30. 28. AdvancedEvaluate lim⁡x→9x−9x−3\lim_{x\to9}\dfrac{x-9}{\sqrt{x}-3}.

  31. 29. AdvancedFind aa and bb so that f(x)=ax+1f(x)=ax+1 for x<3x<3 and f(x)=x2−bf(x)=x^2-b for x≥3x\ge3 is continuous at x=3x=3 with f(3)=7f(3)=7.

  32. 30. AdvancedFor g(x)=x2−x−6x2−9g(x)=\dfrac{x^2-x-6}{x^2-9}, classify the discontinuities at x=3x=3 and x=−3x=-3.