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

2023
Views:

334

Pages:

5
Author:

Shahbaz A.

Q Calculus and analytic geometry shahbaz ahmed August 2024 Introduction to the relevant concepts and skills The Left Hand Limit 1 limx→c− f (x) = L For values of x < c ,as x is getting close to c, the corresponding values of f(x) are getting close to L. The Right Hand Limit 2 limx→c+ f (x) = M For values of x > c ,as x is getting close to c, the corresponding values of f(x) are getting close to M. Criterion for existence of Limit of a Function limx→c f (x) = L ⇔ limx→c− f (x) = limx→c+ f (x) Criterion for Continuity of a Function at a Number A function is said to be continuous at a number ”c” ⇔ (i) f(c) is defined (ii) limx→c f (x) exists, (iii)) limx→c f (x)=f(c) 3 Q If f (x) =  √ √   2x+5− x+7  . x ̸= 2 x−2   k. x = 2 Find value of k so that f is continuous at x=2. Solution limx−→2− √ √ 2x+5− x+7 f (x) = limx−→2− [ .] x−2 Now √ √ 2x+5− x+7 x−2 √ √ √ √ x+7 √2x+5+√x+7 × = 2x+5− x−2 2x+5+ x+7 √ √ 2 ( 2x+5) −( x+7)2 = (x−2)(√2x+5+√x+7) (2x+5)−(x+7) = (x−2)(√2x+5+√x+7) 2x+5−x−7) = (x−2)(√2x+5+√x+7) (x−2) = (x−2)(√2x+5+√x+7) Since x ̸= 2 =⇒ x − 2 ̸= 0 1√ = (√2x+5+ x+7) Putting in limx−→2− f (x) = limx−→2− [ √ √ 2x+5− x+7 .] x−2 4 1√ limx−→2− f (x) = limx−→2− [ (√2x+5+ ] x+7) 1 √ ] limx−→2− f (x) = limx−→2− [ (√2×2+5+ 2+7) limx−→2− f (x) = 1 6 Similarly 1√ limx−→2+ f (x) = limx−→2+ [ (√2x+5+ ] x+7) 1 √ limx−→2+ f (x) = limx−→2+ [ (√2×2+5+ ]= 2+7) Since limx−→2− f (x) = limx−→2+ f (x) = 1 6 =⇒ limx−→2 f (x) = 1 6 For continuity limx−→2 f (x) = f (2) =⇒ 1 6 k= 1 6 =k 5 1 6