Left and right limits
"Left and right limits" is a free, interactive Class 12 Maths simulation for JEE, built on the lesson "Continuity at a point" in Continuity and Differentiability. Come in from the left, then from the right. The hollow rings mark where each side is heading (the left-hand and right-hand limits); the filled dot is the actual value f(c). Continuity needs all three to sit on one spot. F is continuous at c when LHL = RHL = f(c). A jump has unequal one-sided limits; a misplaced dot has equal limits but the wrong value; |x| has a corner yet no break, so it is continuous. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Continuity at a point in Continuity and Differentiability. Change the values and watch what happens.
In JEE Main
42 shifts analysed
48 questions from Continuity and Differentiability across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §5.1–5.2
- Informally, a function is continuous if its graph can be drawn around a point without lifting the pen. The precise test uses limits.
- A real function f is continuous at a point c of its domain when lim f(x) as x → c equals f(c). Three things must hold: f(c) is defined, the limit exists, and the two are equal.
- The limit exists only when the left-hand limit (x approaching c from below) and the right-hand limit (from above) are both finite and equal. So continuity at c means LHL = RHL = f(c).
- Worked example: f(x) = 2x + 3 at x = 1. The limit is 2(1) + 3 = 5 and f(1) = 5, so f is continuous at 1.
- Worked example: f(x) = x³ + 3 for x ≠ 0, with f(0) = 1. Both one-sided limits at 0 are 3, but the value is 1, so f is discontinuous at 0. Redefining f(0) as 3 would repair it.
- Worked example: |x| at 0. From the left |x| = −x → 0, from the right |x| = x → 0, and |0| = 0, so |x| is continuous at 0 even though its graph has a corner there.
- Worked example: the function equal to x + 2 for x ≤ 1 and x − 2 for x > 1. At 1 the left-hand limit is 3 and the right-hand limit is −1, so the graph jumps and f is discontinuous at 1.
- Finding a constant: kx + 1 for x ≤ 5 and 3x − 5 for x > 5 is continuous at 5 only if 5k + 1 equals 15 − 5 = 10, which gives k = 9/5.
Questions
3 answers
- What does the "Left and right limits" simulation show?
- Come in from the left, then from the right. The hollow rings mark where each side is heading (the left-hand and right-hand limits); the filled dot is the actual value f(c). Continuity needs all three to sit on one spot.
- Does JEE Main ask questions from Continuity and Differentiability?
- Yes. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Continuity at a point" in Continuity and Differentiability (Class 12 Maths).
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