MathsMediumClass 12

Limits by Expansion — Taylor series use

Limits & Continuity

JEE Qs

Hard

min

Master the common Maclaurin series expansions and practice expanding functions to just enough terms to resolve indeterminate forms efficiently.

🧮 Key Formulas

e^x = 1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n! + O(x^(n+1))

sin x = x - x^3/3! + x^5/5! - ... + (-1)^n x^(2n+1)/(2n+1)! + O(x^(2n+3))

cos x = 1 - x^2/2! + x^4/4! - ... + (-1)^n x^(2n)/(2n)! + O(x^(2n+2))

ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ... + (-1)^(n-1) x^n/n + O(x^(n+1)) (for |x|<1)

(1+x)^n = 1 + nx + n(n-1)/2! x^2 + n(n-1)(n-2)/3! x^3 + ... + n(n-1)...(n-k+1)/k! x^k + O(x^(k+1)) (for |x|<1)

a^x = e^(x ln a) = 1 + x ln a + (x ln a)^2/2! + ...

tan x = x + x^3/3 + 2x^5/15 + O(x^7)

sin^-1 x = x + x^3/(2*3) + (1*3)/(2*4*5) x^5 + O(x^7) (for |x|<=1)

tan^-1 x = x - x^3/3 + x^5/5 - O(x^7) (for |x|<=1)

Taylor series for f(x) about x=a: f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + ...

Maclaurin series for f(x) (Taylor series about x=0): f(x) = f(0) + f'(0)x/1! + f''(0)x^2/2! + ...

✅ Key Points for JEE

1Maclaurin series (Taylor series about x=0) are generally used for limits as x approaches 0. If the limit is as x approaches 'a', substitute y = x-a, so y approaches 0.
2The primary advantage of series expansion is simplifying complex indeterminate forms (0/0 type) where L'Hopital's Rule becomes tedious due to repeated differentiation.
3Expand each function only up to the degree necessary to resolve the indeterminate form after cancellation. Often, this means one term beyond the lowest power that might cancel.
4When simplifying, group terms by powers of x and cancel common factors from numerator and denominator, focusing on the lowest power terms that remain.
5Be precise with the 'O(x^n)' (order of x to the power n) notation or keep track of sufficient terms to ensure no critical information is lost before simplification.

⚠️ Common Mistakes

✕Incorrectly recalling or applying the series expansions, especially signs and denominators (factorials vs. simple integers).
✕Expanding insufficient number of terms, leading to a remaining 0/0 form or an incorrect limit (e.g., stopping too early and missing the actual lowest power term).
✕Mistaking expansions around x=0 for expansions around other points, or failing to use substitution (y=x-a) when x->a (a != 0).
✕Algebraic errors in simplifying the expanded polynomial expressions.

📝 Practice Questions

See all

Q7. (2x2−3x+5)(3x−1) 2 limx→∞ is equal to : (3x2+5x+4)√(3x+2)x (1) 2 (2) 2e √3e √3 (3) 2 (4) 2e 3√e 3

2025·MCQEasy

Q19.Consider the region R = {(x, y) : x ≤y ≤9 −113 x2, x ≥0}. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is: (1) 730 (2) 625 119 111 (3) 821 (4) 567 123 121

2025·MCQHard

Q11.If limx→∞(( 1−e ) ( e − 1+x )) = α, then the value of 1+loge α equals : (1) e−1 (2) e2 (3) e−2 (4) e

2025·MCQHard

Q7. x2 {sin (k1 + 1)x + sin (k2 −1)x}, x < 0 ⎧ If the function f(x) = 4, x = 0 is continuous at x = 0, then k21 + k22 is ⎨ 2 2+k1x x > 0 x loge ( 2+k2x ), ⎩ equal to (1) 20 (2) 5 (3) 8 (4) 10

2025·Multi conceptHard

Q14. IfI(m, n) = ∫10 xm−1(1 −x)n−1dx, m, (1) I(19, 27) (2) I(9, 1) (3) I(1, 13) (4) I(9, 13)

2025·MCQHard

Q9. Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x) = [x] + |x −2|, −2 < x < 3, is not continuous and not differentiable. Then m + n is equal to : (1) 6 (2) 8 (3) 9 (4) 7

2025·MCQMedium

NCERT Chapters

Class 11 Maths Ch 13: Limits and Derivatives
Class 12 Maths Ch 5: Continuity and Differentiability