ilovepdf_merged (6).pdf). Content covers sections 25.1–25.x.Limit describes the value a function approaches as x approaches a point (not necessarily the value at the point). Continuity means lim x→a f(x) = f(a).
f is continuous at a if: f(a) is defined, limit exists, and they are equal. Polynomials are continuous everywhere; rationals continuous where denominator ≠ 0.
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Limits and Continuity.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: Limit and Continuity
Memorise and apply: x = 2 and at x = 2 ?
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
4 question(s) · Sources: Board-style, Board-style 0/0, Board-style continuity, Board-style standard limit
PYQ1. Evaluate lim_(x→2) (x² − 3x + 1).
Model Answer
1) −1
Explanation
1) Polynomial continuous ⇒ substitute x=2: 4−6+1=−1.
PYQ2. Evaluate lim_(x→3) (x²−9)/(x−3).
Model Answer
1) 6
Explanation
1) Factor (x−3)(x+3)/(x−3) → x+3 → 6 as x→3.
PYQ3. Evaluate lim_(x→0) sin(5x)/x.
Model Answer
1) 5
Explanation
1) sin(5x)/x = 5·sin(5x)/(5x) → 5·1 = 5.
PYQ4. Is f(x)=|x| continuous at x=0? Justify.
Model Answer
1) Yes
Explanation
1) lim_(x→0)|x|=0=f(0). Left and right limits agree with f(0).
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
Find lim_(x→2) (x² − 3x + 1).
1) Polynomial ⇒ continuous; substitute x=2.
2) 4 − 6 + 1 = −1.
Answer: −1
Limits of polynomials are obtained by direct substitution.
Just plug in 2.
No indeterminate form here.
Watch the sign of −3x.
Evaluate lim_(x→3) (x²−9)/(x−3).
1) Factor: (x−3)(x+3)/(x−3) for x≠3.
2) Limit as x→3 of (x+3) = 6.
Answer: 6
Indeterminate 0/0 is resolved by cancelling (x−3).
Difference of squares, cancel, then plug in.
The function is not defined at 3 but limit exists.
Do not cancel and still write x=3 in denominator.
Find lim_(x→0) sin(5x)/(5x).
1) Let u=5x; as x→0, u→0.
2) sin u / u → 1, so limit is 1.
Answer: 1
Standard limit after linear rescaling of the argument.
sin(something small)/(same something) → 1.
If it were sin(5x)/x the limit would be 5.
Keep argument matching top and bottom.
Find lim_(x→0) sin(5x)/x.
1) sin(5x)/x = 5 · (sin(5x)/(5x)) → 5·1 = 5.
Answer: 5
Rewrite to use lim sin u/u=1 with u=5x.
Multiply and divide by 5.
General: lim sin(ax)/x = a as x→0.
Radians assumed.
Is f(x)=|x| continuous at x=0?
1) lim_(x→0)|x|=0 and f(0)=0.
2) Equal ⇒ continuous at 0.
Answer: Yes, continuous at 0
Left and right limits are 0 matching f(0).
Absolute value meets at 0 smoothly in value (though not differentiable).
Continuity ≠ differentiability.
Check both sides for |x|.
f(x)= { x+1 if x<1; 3 if x=1; 2x if x>1 }. Is f continuous at 1?
1) Left limit: 1+1=2.
2) Right limit: 2·1=2.
3) f(1)=3 ≠ 2.
4) Limits exist and agree but ≠ f(1) ⇒ discontinuous at 1 (removable if redefined).
Answer: Discontinuous at x=1 (limit 2 ≠ f(1)=3)
Continuity requires limit equals function value; here 2≠3.
Both sides go to 2, but the plotted point is 3.
Redefining f(1)=2 would make it continuous.
Check all three: L, R, f(a).
Find lim_{x→3} (2x² − 5).
1) Substitute x=3: 2·9 − 5 = 18 − 5 = 13.
Answer: 13
Polynomials are continuous; limit equals value.
Plug in 3.
No indeterminate form.
Check each algebraic step carefully.
Find lim_{x→2} (x² − 4)/(x − 2).
1) (x−2)(x+2)/(x−2) = x+2 for x≠2.
2) Limit as x→2 is 4.
Answer: 4
Remove the 0/0 form by factoring.
Becomes x+2.
Do not substitute before cancelling.
Check each algebraic step carefully.
Find lim_{x→0} (sin 3x)/x.
1) (sin 3x)/x = 3 · (sin 3x)/(3x).
2) As x→0, 3x→0 ⇒ limit = 3 · 1 = 3.
Answer: 3
Standard limit after scaling the argument.
Pull out the factor 3.
Use radians.
Check each algebraic step carefully.
Is f(x)=|x| continuous at x=0?
1) f(0)=0.
2) LHL = RHL = lim_{x→0}|x| = 0 = f(0).
3) Yes, continuous at 0.
Answer: Yes — continuous at x=0
Absolute value is continuous on ℝ.
Both sides approach 0.
Corner does not break continuity.
Check each algebraic step carefully.