Squeeze Theorem
Limits & Continuity
9
JEE Qs
8%
Hard
60
min
Master the technique of finding appropriate bounding functions for oscillatory or complex expressions by utilizing fundamental inequalities and properties of functions.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Squeeze Theorem is primarily used when direct evaluation or standard limit formulas are insufficient, especially for functions involving oscillatory terms like sin(x) or cos(x).
- 2The core idea is to find two simpler functions, g(x) and h(x), that 'squeeze' the target function f(x) from below and above, respectively, and both approach the same limit L at a given point.
- 3The most challenging part is often finding appropriate bounding functions (g(x) and h(x)) by using properties of inequalities and known bounds of functions (e.g., -1 <= sin(x) <= 1).
- 4It's crucial that the limits of the bounding functions exist and are equal; if they are not, the theorem cannot be applied or the bounds are incorrect.
- 5The theorem also applies to limits as x approaches infinity, provided the inequalities hold for sufficiently large x.
⚠️ Common Mistakes
- ✕Incorrectly establishing the bounds for the function, especially when manipulating inequalities involving variables that can change sign (e.g., dividing by x without considering x>0 or x<0).
- ✕Forgetting to verify that both bounding functions (g(x) and h(x)) approach the *same* limit; if they approach different limits, the theorem doesn't apply.
- ✕Applying the theorem without ensuring the inequalities hold in the relevant interval around the limit point (or for sufficiently large/small x if approaching infinity).
📝 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 Maths Ch 13: Limits and Derivatives
- Class 12 Maths Ch 5: Continuity and Differentiability