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

2023
Views:

188

Pages:

6
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 0, θ > 0 √ sin x sin 2θ =1 lim+ √ = lim+ θ−→0 x−→0 x 2θ As √ 2θ is undefined for θ < 0, so limx−→o− sin x x does not exist. Since the limit does not depend on the direction from which x approaches 0 (whether from the positive side or the negative side), the limit remains the same: √ sin x sin 2θ = lim =1 lim √ x−→0 x θ−→0 2θ Intuitive Reasoning: As x gets closer to zero, sin x simultaneously gets closer to values of x in such a way that the ratio shows the behaviour of sin x x sin x x gets closer to 1. The graph below as x get closer to zero . 5 6