University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

442

Pages:

5
Author:

Shahbaz A.

Limit shahbaz ahmed August 2024 1 Reminisce THE SQUEEZE (SANDWICH) THEOREM Let g(x) and h(x) be functions such that g(x) ≤ f (x) ≤ h(x) on a punctured neighborhood of a. If limx−→a g(x) = L and limx−→a h(x) = L ( L ∈ R),then limx−→a f (x) = L A punctured neighborhood of an element x ∈ X is a set which is the relative complement U/{X} 1 2 An important limit Reference to the figure. The line segment AB is a tangent to the circle of radius a at the point B such that OAB is a right angled triangle. Also ∠BOC = x Radians. Where 0 < x < π 2 OB=OC=radii of the circle=a CD is an altitude on OB. Now, in the right angled triangle OAB. AB OB = AB a = tan x AB = a tan x 2 Similarly Area of the circle of radius a with 2π Radians=πa2 2 Area of the sector of a circle of radius a with 1 Radians= πa 2π = a2 2 2 Area of the sector of a circle of radius a with x Radians= a2 x Also, in the right triangle OCD CD OC = CD a = sin x CD = a sin x Area of the triangle OAB= 21 (OB)(AB) = 12 a(a tan x) = 21 a2 tan x Area of the triangle OCB= 12 (OB)(CD) = 21 a(a sin x) = 12 a2 sin x Reference to the figure ⌢ Area of the triangle OCB ≤ Area of the sector of the circle OCB ≤ Area of the triangle OAB 1 2 2 a sin x 2 ≤ a2 x ≤ 21 a2 tan x, 0< x < sin x ≤x ≤ tan x 1 ≤ sinx x ≤ sin x cos x 1 ≤ sinx x ≤ 1 cos x × 1 sin x 1 ≤ limx−→0 sinx x ≤ limx−→0 cos1 x Now limx−→0 cos1 x = 1, since cos 0 = 1 3 π 2 Hence 1 ≤ limx−→0 sinx x ≤ 1 BY THE SQUEEZE (SANDWICH) THEOREM x =1 x−→0 sin x lim Similarly sin x =1 x−→0 x lim 3 Formulas used cos 2x = cos2 x − sin2 x cos2 x = 1 − sin2 x sin2 x = 1 − cos2 x ............................... Evaluate the limit: 1 − cos pθ θ−→0 1 − cos qθ lim Solution 4 1 − cos pθ 1 − cos pθ 1 + cos pθ 1 + cos qθ = lim × × θ−→0 1 − cos qθ θ−→0 1 − cos qθ 1 + cos pθ 1 + cos qθ lim 1 − cos2 pθ 1 1 + cos qθ 1 − cos pθ = lim × × θ−→0 1 − cos2 qθ θ−→0 1 − cos qθ 1 + cos pθ 1 lim 1 − cos pθ 1 − cos2 pθ 1 + cos qθ = lim × θ−→0 1 − cos qθ θ−→0 1 − cos2 qθ 1 + cos pθ lim 1 − cos pθ sin2 pθ 1 + cos qθ lim = lim × θ−→0 1 − cos qθ θ−→0 sin2 qθ 1 + cos pθ sin2 pθ p2 θ2 q 2 θ2 1 + cos qθ 1 − cos pθ = lim 2 2 × 2 2 × θ−→0 p θ θ−→0 1 − cos qθ q θ sin2 qθ 1 + cos pθ lim p2 sin pθ 2 1 − cos pθ qθ 2 1 + cos qθ lim = 2 [ lim ] × [ lim ] × [ lim ] θ−→0 1 − cos qθ θ−→0 sin qθ θ−→0 1 + cos pθ q θ−→0 pθ p2 2 1 − cos pθ lim = 2 [1] × [1]2 × [1] θ−→0 1 − cos qθ q 1 − cos pθ p2 lim = 2 θ−→0 1 − cos qθ q 5