☰ Contents · Mathematics

The idea of a limit

Lessons 2 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
2

The idea of a limit

Textbook, Part 1: pp. 12–15
GoalUnderstand the limit of a function (approach from the left and from the right) and compute simple limits by substitution and simplification.
New words
limit · limitapproach from the left / right · chapdan / o‘ngdan yaqinlashishindeterminate form 0/0 · aniqmaslik 0/0point of discontinuity · uzilish nuqtasi
Explanation

If, as x approaches a (with x ≠ a), the values f(x) approach a number A, then A is called the limit of f at a, written lim f(x) = A as x → a. The approach must give the same number from the left (x < a) and from the right (x > a); if the two sides give different numbers, the limit does not exist. The limit does not depend on the value f(a): the function may be undefined at a or have a different value there; only the values at points near a matter. For continuous expressions the limit is found by plain substitution. If substitution gives 0/0, use x ≠ a: factorise, cancel, simplify, and then take the limit.

Worked examples
lim (x² + 3x + 2)/(x + 2) as x → −2. Substitution gives 0/0. For x ≠ −2 factorise the numerator: x² + 3x + 2 = (x + 1)(x + 2) and cancel to get x + 1. The limit is −2 + 1 = −1.
For f(x) = |x|/x at x = 0: for x > 0, f(x) = 1 and for x < 0, f(x) = −1. The right-hand limit is 1, the left-hand limit is −1; they differ, so the limit as x → 0 does not exist. Another example: if f(x) = x + 1 for x ≠ 2 and f(2) = 10, then lim f(x) = 3 as x → 2, because f(2) does not matter.
Class activity

“Is there a limit?”: the teacher puts 4–5 different functions on the board (piecewise, fractional, |x|/x). In pairs, students tabulate x-values approaching a from the left and from the right, decide whether a limit exists, and check each other’s tables.

Practice
1
Compute lim (3x² − 5x) as x → 4.
2
Find lim (x² − 1)/(x + 1) as x → −1.
3
Compute lim ((3 + h)² − 9)/h as h → 0.
4
Why may we cancel (x − 3) when finding a limit as x → 3, even though the denominator is zero at x = 3?