Lesson 1 of 13 · 11 min
Continuity at a point
NCERT §5.1–5.2
Riya rides her scooter to college. Her odometer never skips a number: to get from 5 km to 6 km it passes through every reading in between. Her phone's tracking app, one day, showed her jumping 4 km backwards in an instant. Which of the two can be trusted?
The story this chapter follows: Riya's scooter ride
The lesson in notes
In short
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.