Standard Limits — sinx/x, (eˣ-1)/x, (aˣ-1)/x, (1+1/x)ˣ
Limits & Continuity
9
JEE Qs
8%
Hard
75
min
Master the art of transforming complex limit expressions into standard forms by algebraic manipulation and substitution, always verifying that arguments match and the variable approaches the correct value.
🧮 Key Formulas
✅ Key Points for JEE
- 1For trigonometric and exponential/logarithmic limits, the variable must approach zero, and the argument of the function must be identical to the denominator (e.g., sin(f(x))/f(x) where f(x) -> 0).
- 2For the (1+1/x)^x type limits, the expression must be of the indeterminate form 1^infinity, where the base tends to 1 and the exponent tends to infinity.
- 3Complex limit problems often require algebraic manipulation (e.g., factorization, rationalization, substitution) to transform the expression into one or more standard limit forms.
- 4These standard limits can be derived using series expansions (Maclaurin series) or L'Hopital's Rule, providing a deeper understanding and alternative verification methods.
- 5The limit lim (x->0) ( (1+x)^n - 1 ) / x = n is a generalization often useful when dealing with binomial expansions for fractional or negative powers.
⚠️ Common Mistakes
- ✕Not ensuring the argument of the function and the denominator are exactly the same (e.g., treating sin(2x)/x as 1 directly without multiplying by 2/2).
- ✕Incorrectly applying standard limits when the variable does not tend to the required value (e.g., using lim (x->infinity) sinx/x = 1 instead of lim (x->0) sinx/x = 1).
- ✕Failing to recognize the indeterminate form 1^infinity or misapplying the 'e' form, especially when the base or exponent is complex.
- ✕Forgetting the 'ln a' term in lim (x->0) (a^x-1)/x, or using log base 10 instead of natural log (ln).
- ✕Applying L'Hopital's rule prematurely or without verifying the indeterminate form, when simple algebraic manipulation or standard limits would be faster.
📝 Practice Questions
See allQ7. (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
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
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
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
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)
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
NCERT Chapters
- Class 11 Mathematics Ch 13: Limits and Derivatives
- Class 12 Mathematics Ch 5: Continuity and Differentiability